f 


HIGHER  ALGEBRA 


BY 

G.   A.   WENTWORTH, 

PROFESSOR  OF  MATHEMATICS   IN  PHILLIPS   EXETER  ACADEMY. 


3XXC 


.^ 

< 

BOSTON,   U.S.A.: 

BLISHED   BY   GINK   & 

COMPANY. 

1891. 

\^'^i 


\ 


Entered  according  to  Act  of  Congress,  in  the  year  1891,  by 

G.  A.  WENTWORTH, 
in  the  OflOice  of  the  Librarian  of  Congress,  at  Washington. 


All  Rights  Reserved. 


Typography  by  J.  S.  Cushing  &  Co.,  Boston,  U.S.A. 


Presswork  by  Ginn  &  Co.,  Boston,  U.S.A. 


PREFACE. 


rriHIS  work  is  intended  to  give  in  one  book  a  thorough  prepara- 
tory course  for  Colleges  and  Scientific  Schools,  and  in  addition 
a  sufficiently  full  treatment  of  the  subjects  usually  read  by  students 
in  general  in  such  institutions.  In  short,  it  provides  a  course 
parallel  to  the  course  covered  by  the  author's  School  and  College 
Algebras  together.  The  elementary  part  is  as  full  as  the  School 
Algebra;  the  advanced  part,  however,  is  briefer  than  the  College 
Algebra.  The  book  is  substantially  equivalent  to  the  author's 
Complete  Algebra,  but  is  greatly  superior  to  that  work  in  the 
arrangement  of  topics  and  in  the  methods  of  presenting  them. 

Preparatory  Schools  and  Academies,  if  their  pupils  have  had  a 
thorough  drill  in  Arithmetic  before  they  begin  the  study  of  Algebra, 
will  find  this  book  specially  suited  to  their  needs.  The  brighter 
boys  of  the  class  can  read  the  advanced  chapters  while  the  duller 
boys  review  the  elementary  chapters. 

Colleges  and  Scientific  Schools,  if  their  pupils,  owing  to  lack 
of  previous  drill,  have  to  review  carefully  the  preparatory  work 
of  the  schools  before  entering  upop  the  college  work  proper,  will  find 
in  this  a  convenient  and  sufficiently  full  book  for  their  requirements. 

Answers  to  the  problems  are  bound  separately,  in  paper  covers, 
and  will  be  furnished  free  for  pupils  when  teachers  apply  to  the  pub- 
lishers for  them. 

Any  corrections   or  suggestions    relating    to  the  work   will  be 

thankfully  received. 

G.  A.  WENTWORTH. 

Phillips  Exeter  Academy, 
Exeter,  N.H.,  May,  1891. 


270247 


CONTENTS. 


Chaptsr  Paqe 

I.  Definitions 1 

II.  Addition  and  Subtraction 12 

III.  Multiplication 26 

IV.  Division 39 

V.  Simple  Equations 48 

VI.  Multiplication  and  Division 61 

VII.  Factors 70 

VIII.  Common  Factors  and  Multiples 93 

IX.  Fractions 109 

X.  Fractional  Equations 134 

XI.  Simultaneous  Equations  of  the-JFirst  Degree  154 

XII.  Problems  involving  Two  Unknown  Numbers  .  169 

XIII.  Simple  Indeterminate  Equations 186 

XIV.  Inequalities » 192 

XV.  Involution  and  Evolution 194 

XVI.  Theory  of  Exponents 207 

XVII.  Radical  Expressions 214 

XVIII.  Imaginary  Expressions 229 

XIX.  Quadratic  Equations .  234 

XX.  Simultaneous  Quadratic  Equations   ....  260 

XXI.  Properties  of  Quadratics 271 

XXII.  Ratio,  Proportion,  and  Variation     ....  277 


VI  CONTENTS. 

Chaptbr  Pias 

XXIII.  Peogressions 295 

XXIV.  Indeterminate  Coefficients  .     .     .     .     .     .     .  312 

XXV.     Binomial  Theorem 319 

XXVI.    Common  Logarithms 331 

XXVII.    Interest  and  Annuities 346 

XXVIII.    Choice 356 

XXIX.    Chance 377 

XXX.    Continued  Fractions 381 

XXXI.    Scales  of  Notation 392 

XXXII.    Theory  of  Numbers 397 

XXXIII.  Variables  and  Limits 403 

XXXIV.  Series 412 

XXXV.  General  Properties  of  Equations     ....  441 

.  XXXVI.    Numerical  Equations 473 

XXXVII.    Determinants 499 

XXXVIII.    Complex  Numbers 515 


HIGHER  ALGEBRA. 


CHAPTER  I. 
DEFINITIONS. 

1.  Units.  In  counting  separate  objects  the  standards  by 
which  we  count  are  called  units ;  and  in  measuring  contin- 
uous magnitudes  the  standards  by  which  we  measure  are 
called  units. 

Thus,  in  counting  the  boys  in  a  school,  the  unit  is  a  boy  ;  in  sell- 
ing eggs  by  the  dozen,  the  unit  is  a  dozen  eggs  ;  in  selling  bricks  by 
the  thousand,  the  unit  is  a  thousand  bricks  ;  in  measuring  short  dis- 
tances, the  unit  is  an  inch,  a  foot,  or  a  yard ;  in  measuring  long 
distances,  the  unit  is  a  rod  or  a  mile. 

2.  Numbers.  Repetitions  of  the  unit  are  expressed  by 
numbers.  If  a  man,  in  sawing  logs  into  boards,  wishes  to 
keep  a  count  of  the  logs,  he  makes  a  straight  mark  for 
every  log  sawed,  and  his  record  at  different  times  will  be 
as  follows : 

/    //    ///    ////    fuj     ruj  I 
tw  II    iw  III    mi  nil    nu  iHJ 

These  representative  groups  are  named  one,  two,  th^ee, 
four,  five,  six,  seven,  eight,  nine,  ten,  etc.,  and  are  known 
collectively  under  the  general  name  of  numbers.  It  is 
obvious  that  these  representative  groups  will  have  the 
same  meaning,  whatever  the  nature  of  the  unit  counted. 


2  ALGEBRA. 

3.  Quantities.  The  word  "  quantity "  (from  the  Latin 
quantus,  how  much)  implies  both  a  unit  and  a  number. 
Thus,  if  we  inquire  how  much  wheat  a  bin  will  hold,  we 
mean  how  many  bushels  of  wheat  it  will  hold. 

4.  Number-Symbols  in  Arithmetic.  Instead  of  groups  of 
straight  marks,  we  use  in  Arithmetic  the  arbitrary  sym- 
bols 1,  2,  3,  4,  5,  6,  7,  8,  9,  called  figures,  for  the  numbers 
one,  two,  three,  four,  five,  six,  seven,  eight,  nine. 

The  next  number,  ten,  is  indicated  by  writing  the  figure 
1  in  a  different  position,  so  that  it  shall  signify  not  one,  but 
te7i.  This  change  of  position  is  effected  by  introducing  a 
new  symbol,  0,  called  nought  or  zero,  and  signifying  none. 
Thus,  in  the  symbol  10,  the  figure  1,  occupying  the  second 
place  from  the  right,  signifies  a  collection  of  ten  things,  and 
the  zero  signifies  that  there  are  no  single  things  over.  The 
symbol  11  denotes  a  collection  of  ten  things  and  one  thing 
besides.  All  succeeding  numbers  up  to  the  number  con- 
sisting of  10  tens  are  expressed  by  writing  the  figure  for 
the  number  of  tens  they  contain  in  the  second  place  from 
the  right,  and  the  figure  for  the  number  of  units  besides 
in  the  first  place.  The  number  consisting  of  10  tens  is 
called  a  hundred,  and  the  hundreds  of  a  number  are  written 
in  the  third  place  from  the  right.  The  number  consisting 
of  10  hundreds  is  called  a  thousand,  and  the  thousands  are 
written  in  the  fourth  place  from  the  right;  and  so  on. 

5.  Number-Symbols  in  Algebra.  Algebra,  like  Arithmetic, 
treats  of  numbers,  and  employs  the  letters  of  the  alphabet  in 
addition  to  the  figures  of  Arithmetic  to  represent  numbers. 
The  letters  of  the  alphabet  are  used  as  general  symbols  of 
numbers  to  which  any  particular  values  may  be  assigned. 
In  any  problem,  however,  a  letter  must  be  supposed  to 
have  the  same  particular  value  throughout  the  investiga- 
tion or  discussion  of  the  problem. 


DEFINITIONS.  3 

These  general  symbols  are  of  great  advantage  in  investi- 
gating and  stating  general  laws ;  in  exhibiting  the  actual 
method  in  which  a  number  is  made  up ;  and  in  represent- 
ing unknown  numbers  which  are  to  be  discovered  from 
their  relations  to  known  numbers. 

6.  Principal  Signs  of  Operations.  The  signs  of  the  funda- 
mental operations  are  the  same  in  Algebra  as  iri  Arith- 
metic, and  are 

The  sign  +  (read  plus),  the  sign  of  addition. 

The  sign  —  (read  minus),  the  sign  of  subtraction. 

The  sign  X  (read  times  or  into),  the  sign  of  multiplication. 

The  sign  -=-  (read  divided  hy),  the  sign  of  division. 

The  multiplier  is  sometimes  written  after  the  sign,  and 
then  the  sign  is  read  multiplied  hy. 

The  operation  of  division  is  often  indicated  by  writing 
the  dividend  over  the  divisor  with  a  line  between  them. 
Thus  f  means  the  same  as  8  h-  4. 

7.  Signs  of  Kelation.  The  signs  of  relation  are  =,>,<, 
which  stand  for  the  words,  "is  equal  to,"  "is  greater  than," 
and  "  is  less  than,"  respectively. 

8.  Signs  of  Aggregation.     The  signs  of  aggregation  are 

the  bar,  | ;  the  vinculum, ;  the  parenthesis,  (  ) ;  the 

bracket,  [  ]  ;  and  the  brace,  { | .  Thus,  each  of  the  expres- 
sions, ,?,  a-\-b,  (a+i),  [a+^],  \a-{-h\,  signifies  that 
a-\-h  is  to  be  treated  as  a  single  number. 

9.  Signs  of  Oontinuation.  The  signs  of  continuation  are 
dots, ,  or  dashes, ,  and  are  read,  "  and  so  on." 

10.  Sign  of  Deduction.  The  sign  of  deduction  is  .'.,  and 
is  read,  "  hence  "  or  "  therefore." 


ALGEBRA. 


11.  The  Natural  Series  of  Numbers.  Beginning  with  the 
number  one,  each  succeeding  number  is  obtained  by  put- 
ting one  more  with  the  preceding  number. 


7    etc. 


Thus,  if  from  a  given  point  marked  0,  we  draw  a  straight 
line  to  the  right,  and  beginning  from  this  point  lay  oif 
units  of  length,  the  successive  repetitions  of  the  unit  will 
be  denoted  by  the  natural  series  of  numbers,  1,  2,  3,  4,  etc. 


12.  Positive  and  Negative  Numbers.  There  are  quantities 
which  stand  to  each  other  in  such  opposite  relations  that, 
when  we  combine  them,  they  cancel  each  other  entirely  or 
in  part.  Thus,  six  dollars  gain  and  six  dollars  loss  just 
cancel  each  other;  but  ten  dollars  gain  and  six  dollars 
loss  cancel  each  other  only  in  part.  For  the  six  dollars 
loss  will  cancel  six  dollars  of  the  gain  and  leave  four 
dollars  gain. 

An  opposition  of  this  kind  exists  in  assets  and  debts,  in 
motion  forwards  and  motion  bachiuards,  in  motion  to  the 
right  and  motion  to  the  left,  in  the  degrees  above  and  the 
degrees  below  zero  on  a  thermometer. 

From  this  relation  of  quantities  a  question  often  arises 
which  is  not  considered  in  Arithmetic ;  namely,  the  sub- 
tracting of  a  greater  number  from  a  smaller.  This  cannot 
be  done  in  Arithmetic,  for  the  real  nature  of  subtraction 
consists  in  counting  backwards,  along  the  natural  series  of 
numbers.  If  we  wish  to  subtract  four  from  six,  we  start 
at  six  in  the  natural  series,  count  four  units  backwards,  and 
arrive  at  two,  the  difference  sought.  If  we  subtract  six 
from  six,  we  start  at  six  in  the  natural  series,  count  six 
units  backwards,  and  arrive  at  zero.     If  we  try  to  subtract 


DEFINITIONS.  5 

nine  from  six,  we  cannot  do  it,  because,  when  we  have 
counted  backwards  as  far  as  zero,  the  natural  senes  of 
numhef)^s  comes  to  an  end. 

13.  In  order  to  subtract  a  greater  number  from  a  smaller, 
it  is  necessary  to  assume  a  new  series  of  numbers,  begin- 
ning at  zero  and  extending  to  the  left  of  zero.  The  series 
to  the  left  of  zero  must  ascend  from  zero  by  the  repetitions 
of  the  unit,  precisely  like  the  natural  series  to  the  right  of 
zero  ;  and  the  opposition  between  the  right-hand  series  and 
the  left-hand  series  must  be  clearly  marked.  This  opposi- 
tion is  indicated  by  calling  every  number  in  the  right-hand 
series  a  positive  number,  and  prefixing  to  it,  when  written, 
the  sign  + ;  and  by  calling  every  number  in  the  left-hand 
series  a  negative  number,  and  prefixing  to  it  the  sign  — . 
The  two  series  of  numbers  will  be  written  thus : 

-4,-3,-2,-1,   0,    +1,  +2,  +3,  +4,  

I I ! I I        I        I J I 

If,  now,  we  wish  to  subtract  9  from  6,  we  begin  at  6  in 
the  positive  series,  count  nine  units  in  the  negative  direction 
(to  the  left),  and  arrive  at  —  3  in  the  negative  series.  That 
is,  6-9=-3. 

The  result  obtained  by  subtracting  a  greater  number 
from  a  less,  when  both  are  positive,  is  always  a  negative 
number. 

If  a  and  h  represent  any  two  numbers  of  the  positive 
series,  the  expression  a—h  will  denote  a  positive  number 
when  a  is  greater  than  h  ;  will  be  equal  to  zero  when  a  is 
equal  to  h ;  will  denote  a  negative  number  when  a  is  less 
than  h. 

If  we  wish  to  add  9  to  —6,  we  begin  at  —6,  in  the 
negative  series,  count  nine  units  in  the  positive  direction 
(to  the  right),  and  arrive  at  -f-3,  in  the  positive  series. 


6  ALGEBRA. 

We  may  illustrate  the  use  of  positive  and  negative 
numbers  as  follows : 

-5  0  8  20 

1 1 1 1 

DA  C 

Suppose  a  person  starting  at  A  walks  20  feet  to  the 
right  of  A,  and  then  returns  12  feet,  where  will  he  be? 
Answer:  at  C,  a  point  8  feet  to  the  right  of  A.  That  is, 
20  feet  — 12  feet  equals  8  feet. 

Again,  suppose  he  walks  from  A  to  the  right  20  feet, 
and  then  returns  25  feet,  where  will  he  now  be?  Answer: 
at  D,  a  point  5  feet  to  the  left  of  A.  That  is,  if  we  con- 
sider distances  measured  in  feet  to  the  left  of  A  as  expressed 
by  a  negative  series  of  numbers,  beginning  at  A,  20  feet— 
25  feet  equals  —  5  feet.  Hence,  the  phrase,  5  feet  to  the  left 
of  J.,  is  now  expressed  by  the  negative  quantity,  —  5  feet. 

Remark.,  In  Arithmetic,  if  the  things  counted  are  whole  units, 
the  numbers  which  count  them  are  called  whole  nmnbers,  integral  num- 
bers, or  integers,  where  the  adjective  is  transferred  from  the  things 
counted  to  the  numbers  which  count  them.  But  if  the  things  counted 
are  only  parts  of  units,  the  numbers  which  count  them  are  called 
fractional  numljers,  or  simply  fractions,  where  again  the  adjective  is 
transferred  from  the  things  counted  to  the  numbers  which  count 
them. 

In  Algebra,  if  the  units  counted  are  negative,  the  numbers  which 
count  them  are  called  negative  numbers,  where  the  adjective  which 
defines  the  nature  of  the  units  counted  is  transferred  to  the  numbers 
that  count  them. 

14.  Numbers  with  the  sign  +  oi*  —  are  called  algebraic 
numbers.  They  are  unknown  in  Arithmetic,  but  play  a 
very  important  part  in  Algebra.  Numbers  not  affected 
by  the  signs  +  or  —  are  called  absolute  numbers. 

Every  algebraic  number,  as  +4  or  —4,  consists  of  a 
sign  +  or  —  and  the  absolute  value  of  the  number ;  in  this 
case  4.     The  sign  shows  whether  the  number  belongs  to 


DEFINITIONS.  7 

the  positive  or  negative  series  of  numbers ;  the  absolute 
value  shows  the  place  the  number  has  in  the  positive  or 
negative  series. 

When  no  sign  stands  before  a  number,  the  sign  +  is 
always  understood;  thus,  4  means  the  same  as  +4,  a 
means  the  same  as  -\-a.     But  the  sign  —  is  never  omitted. 

Two  numbers  which  have,  one  the  sign  -f  and  the  other 
the  sign  — ,  are  said  to  have  unlike  signs. 

Two  numbers  which  have  the  same  absolute  values,  but 
unlike  signs,  always  cancel  each  other  when  combined ; 
thus,  +4-4=0,  +a-a  =  0. 

15.  Meaning  of  the  Signs.  The  use  of  the  signs  +  and  — , 
to  indicate  addition  and  subtraction,  must  be  carefully  dis- 
tinguished from  their  use  to  indicate  in  which  series,  the 
positive  or  the  negative,  a  given  number  belongs.  In  the 
first  sense,  they  are  signs  of  operations,  and  are  common  to 
both  Arithmetic  and  Algebra.  In  the  second  sense,  they 
are  signs  of  opposition,  and  are  employed  in  Algebra 
alone. 

16,  Factors.  When  a  number  consists  of  the  product  of 
two  or  more  numbers,  each  of  these  numbers  is  called  a 
factor  of  the  product. 

The  sign  X  is  generally  omitted  between  a  figure  and  a 
letter,  or  between  letters ;  thus,  instead  of  63  X  a  X  5,  we 
write  63  a5  ;  instead  of  a  X  6  X  c,  we  write  ahc. 

The  expression  ahc  must  not  be  confounded  with  a-\-h-\-c. 

If  a  =  2,    6  =  3,    c  =  4, 

then  a6c  =  2  X  3  X  4  =  24  ; 

and  a  -f  Z>  H-  c  =  2  +  3  +  4  =  9. 

Factors  expressed  by  letters  are  called  literal  factors; 
factors  expressed  by  figures  are  called  numerical  factors. 


8  ALGEBRA. 

17.  Coefficients.  A  known  factor  of  a  product  which  is 
prefixed  to  another  factor,  to  show  how  many  times  that 
factor  is  taken,  is  called  a  coefficient.  Thus,  in  *J  a,  7  is  the 
coefficient  of  c ;  in  7  ax,  7  is  the  coefficient  of  ax,  or,  if  a  is 
known,  7  a  is  the  coefficient  of  x. 

By  coefficient,  we  generally  mean  the  numerical  coeffi- 
cient with  its  sign.  If  no  numerical  coefficient  is  written, 
1  is  understood.     Thus,  ax  means  the  same  as  1  ax. 

18.  Powers.  A  product  consisting  of  two  or  more  equal 
factors  is  called  a  power  of  that  factor. 

19.  Indices  or  Exponents.  An  index  or  exponent  is  a 
number-symbol  written  at  the  right  of,  and  a  little  above,  a 
number. 

If  the  index  is  a  positive  integral  number,  it  shows  the 
number  of  times  the  given  number  is  taken  as  a  factor. 

Thus,  a},  or  simply  a,  denotes  that  a  is  taken  once  as  a  factor ;  a^ 
denotes  that  a  is  taken  twice  as  a  factor  ;  o?  denotes  that  a  is  taken 
three  times  as  a  factor ;  and  a"  denotes  that  a  is  taken  n  times  as  a 
factor.  These  are  read :  the  first  power  of  a ;  the  second  power  of  a ; 
the  third  power  of  a  ;  the  nth  power  of  a. 

o?  is  written  instead  of  aaa. 

a^  is  written  instead  of  aaa,  etc.,  repeated  n  times. 

The  meaning  of  coefficient  and  exponent  must  be  care- 
fully distinguished.     Thus, 

a^  =  axaxaxa. 

If  a -3,  4a  =  3 +  3 +  3  +  3  =  12. 

a*  =  3  X  3  X  3  X  3  =  81. 

The  second  power  of  a  number  is  generally  called  the  square  of 
that  number;  thus,  a^  is  called  the  square  of  a,  because  if  a  denotes 
the  number  of  units  of  length  in  the  side  of  a  square,  a^  denotes  the 
number  of  units  of  surface  in  the  square.  The  third  power  of  a  num- 
ber is  generally  called  the  cuhe  of  that  number  ;  thus,  a^  is  called  the 
cube  of  a,  because  if  a  denotes  the  number  of  units  of  length  in  the 
edge  of  a  cube,  a^  denotes  the  number  of  units  of  volume  in  the  cube. 


DEFINITIONS.  9 

20.  Eoots.  The  root  of  a  number  is  one  of  the  equal  fac- 
tors of  that  number  ;  the  square  root  of  a  number  is  one  of 
the  two  equal  factors  of  that  number ;  the  cube  root  of  a 
number  is  one  of  the  three  equal  factors  of  that  number ; 
and  so  on.  The  sign  ^,  called  the  radical  sign,  indicates 
that  a  root  is  to  be  found.  Thus,  V4,  or  V4,  means  that 
the  square  root  of  4  is  to  be  taken  ;  VS  means  that  the 
cube  root  of  8  is  to  be  taken ;  and  so  on.  The  figure  writ- 
ten above  the  radical  sign  is  called  the  index  of  the  root. 

21.  Algebraic  Expressions.  An  algebraic  expression  is 
a  number  written  with  algebraic  symbols;  an  algebraic 
expression  consists  of  one  symbol,  or  of  several  symbols 
connected  by  signs  of  operation. 

A  term  is  an  algebraic  expression  the  parts  of  which  are 
not  separated  by  the  sign  of  addition  or  subtraction.  Thus, 
3a5,  5a:y,  3aZ>  X  bxy,  ^ab  ^bxy  are  terms. 

A  monomial  or  simple  expression  is  an  expression  with  but 
one  term. 

A  polynomial  or  compound  expression  is  an  expression  of 
two  or  more  terms.  A  binomial  is  a  polynomial  of  two 
terms  ;  a  trinomial,  a  polynomial  of  three  terms. 

Like  terms  or  similar  terms  are  terms  which  have  the  same 
letters,  and  the  corresponding  letters  affected  by  the  same 
exponents.     Thus,  7  a?cx^  and  —  5  a^coi^  are  like  terms. 

22.  The  degree  of  a  term  is  the  sum  of  the  exponents  of 
its  literal  factors.     Thus,  ^xy  is  of  the  second  degree. 

A  polynomial  is  said  to  be  homogeneous  when  all  its 
terms  are  of  the  same  degree.  Thus,  1  x^  —  b  xhj -\- xyz  is 
homogeneous  of  the  third  degree. 

A  polynomial  is  said  to  be  arranged  according  to  the 
powers  of  some  letter  when  the  exponents  of  that  letter 
either  descend  or  ascend  in  order  of  magnitude. 


10  ALGEBRA. 

23.    The  value  of  an  algebraic  expression  is  the  number 
which  the  expression  represents. 

Exercise  1. 

If  a=l,  5  =  2,  c  =  3,   c?  =  4,  e  =  5,f=0,    find   the 
values  of  the  following  expressions  : 

1     n      io7io         o  A     4:ac  ,  8bc      bed 

1.  9a  + 26  +  3c  — 2/.  4.    — — \-— 

ode 

2.  4e-3a-35  +  5c.  5.    le  +  bcd-^^- 

3 .  8ahc  —  hcd+^cde  —  def.     6 .    abc^  +  bcd"^  —  deo?  -{-J^. 

7.    e*  +  6e'^6^  +  5*-4e'^>-4e5l 
„     8a^  +  36^  ,  4c^  +  66^      c^  +  c^^ 

o. 


10. 


Note.  In  finding  the  value  of  a  compound  expression  the  opera- 
tions indicated  for  each  term  must  be  performed  before  the  operation 
indicated  by  the  sign  prefixed  to  the  terra.  Indicated  divisions 
should  be  written  in  the  fractional  form,  and  the  sign  X  should  be 
omitted  between  a  figure  and  a  letter,  or  between  two  letters.    Thus, 

(6  -  c)  -4-  2  X  c  +  26  should  be  written  ^^^  +  26. 

2i  c 

Simplify  the  following  expressions  : 

13.  100  +  80-^4.  15.    25  +  5x4-10-^-5. 

14.  75-25x2.  16.   24-5x4^10  +  3. 

17.    (24  -  5)  X  (4 -^10 +  3). 
If  a=^2,b  =  \0,  x  =  ?>,  y  =  b,  find  the  value  of 

18.  a:?/  +  4ax2.  20.    3a:  +  7y -^  7  +  a  X  y. 

19.  xy-lbb-^b.      '  21.    6b -8y  ^2j/ X  b -2b. 


a'b' 

'      c'  -  b'              e' 

d'^ 
b^' 

11.         ^'  +  ^'     . 
b-'  +  d'-bd 

e'  +  b'' 

e^-d^ 

c'  -  h^ 

"'    e'  +  ed  +  d' 

DEFINITIONS.  11 

22.  {<oh-Sy)^2yxh-\-2h, 

23.  &h  —  {Sy-^2y)xb  —  2h. 

24.  Qh^{b-y)-Zx-\-bxy-~lQa. 

Exercise  2. 

1.  Express  the  sum  of  a  and  b.      > 

2.  Express  the  double  of  :r. 

3.  By  how  much  is  a  greater  than  5? 

4.  If  :r  is  a  whole  number,  find  the  next  number  above 

it. 

6.    Write  five  numbers  in  order  of  magnitude,  so  that  x 
shall  be  the  middle  number. 

6.  What  is  the  sum  of  a;  +  ^  +  ^  + writtep  a  times? 

7.  If  the  product  is  xy  and  the  multiplier  x,  what  is  the 

multiplicand  ? 

8.  A  man  who  has  a  dollars  spends  b  dollars;  how  many- 

dollars  has  he  left? 

9.  A  regiment  of  men  can  be  drawn  up  in  a  ranks  of  b 

men  each,  and  there  are  c  men  over;  of  how  many- 
men  does  the  regiment  consist? 

10.  Write,  the  sum  of  x  and  y  divided  by  c  is  equal  to  the 

product  of  a,  b,  and  m,  minus  six  times  c,  plus  the 
quotient  of  a  divided  by  the  sum  of  x  and  y. 

11.  Write,  six  times  the  square  of  n,  divided  by  m  minus 

a,  plus  five  b  into  the  expression  c  plus  d  minus  a. 

12.  Write,  four  times  the  fourth  power  of  a,  diminished  by 

five  times  the  square  of  a  into  the  square  of  b,  and 
increased  by  three  times  the  fourth  power  of  b. 


CHAPTER   II. 


i^K 


A.. 


ADDITION   AifD   SUBTRACTION. 

Integral  Expressions. 

24.  If  an  algebraic  expression  contains  only  integral 
forms,  that  is,  contains  no  letter  in  the  denominator  of 
any  of  its  terms,  it  is  called  an  integral  expression.  Thus, 
x^  -\- 1  cx^  —  c^  —  b c"^ X  and  iax  —  ^hcy  are  integral  expres- 

sions,  but is  a  fractional  expression. 

o?  —  ab-\-  IP" 

An  integral  expression  may  have  for  some  values  of  the  letters  a 
fractional  value,  and  a  fractional  expression  an  integral  value.  If, 
for  instance,  a  stands  for  f  and  h  for  },  the  integral  expression 
2  a  —  5  6  stands  for  |  —  |  =  |- ;  and  the  fractional  expression  —  stands 

o  0 

for  if^-  -J- 1  =  5.  Integral  and  fractional  expressions,  therefore,  are  so 
named  on  account  of  the  form  of  the  expressions,  and  with  no  refer- 
ence whatever  to  the  numerical  value  of  the  expressions  when  defi- 
nite numbers  are  put  in  place  of  the  letters. 

25.  Definition  of  Addition.  The  process  of  finding  the 
result  when  two  or  more  numbers  are  taken  together  is 
called  addition,  and  the  result  is  called  the  sum. 

26.  Definition  of  Subtraction.  The  process  of  finding  the 
result  when  one  number  is  taken  from  another  is  called 
subtraction,  and  the  result  is  called  the  difference  or  re- 
mainder. 

The  number  taken  away  is  called  the  subtrahend;  the 
number  from  which  the  subtrahend  is  taken  is  called  the 
minuend. 


ADDITION    AND    SUBTRACTION.  13 

In  practice  the  difference  is  found  by  discovering  the 
number  which  must  be  added  to  the  subtrahend  to  give 
the  minuend.    Hence  the  general  definition  of  subtraction  is 

The  operation  of  finding  from  two  given  numbers,  called 
minuend  and  subtrahend,  a  third  number,  called  difference, 
which  added  to  the  subtrahend  will  give  the  minuend. 

27.  Parentheses  for  Algebraic  Numbers.  An  algebraic 
number  which  is  to  be  added  or  subtracted  is  often  inclosed 
in  a  parenthesis,  in  order  that  the  signs  +  and  — ,  which 
are  used  to  distinguish  positive  and  negative  numbers,  may 
not  be  confounded  with  the  +  and  —  signs  that  denote  the 
operations  of  addition  and  subtraction.  Thus  -f  4  +  (—  3) 
expresses  the  sum,  and  +  4  —  (—  3)  expresses  the  difier- 
ence,  of  the  numbers  -\-  4  and  —  3. 

28.  Addition  of  Algebraic  Numbers.  In  order  to  add  two 
algebraic  numbers,  we  begin  at  the  place  in  the  series  which 
the  first  number  occupies  and  count,  in  the  direction  indi- 
cated by  the  sign  of  the  second  number,  as  many  units  as 
there  are  in  the  absolute  value  of  the  second  number. 

The  sum  of  +  4  +  (+  3)  is  found  by  counting  from  +  4  three  units 
in  the  positive  direction;  that  is,  to  the  right,  and  is,  therefore,  +  7. 

The  sum  of  +  4  +  (—  3)  is  found  by  counting  from  +  4  three  units 
in  the  negative  direction;  that  is,  to  the  left,  and  is,  therefore,  +  1. 

-5    -4-3-2-1      0+1+2+3+4+5    +6 

The  sum  of  —  4  +  (+  3)  is  found  by  counting  from  —  4  three  units 
in  the  positive  direction,  and  is,  therefore,  —  1. 

The  sum  of  —  4  +  (—  3)  is  found  by  counting  from  —  4  three  units 
in  the  negative  direction,  and  is,  therefore,  —  7. 

Hence  to  add  two  or  more  algebraic  numbers,  we  have 
the  following  rules : 

I.  If  the  numbers  have  UTce  signs.  ^  Find  the  sum  of  their 
absolute  values,  and  prefix  the  common  sign  to  the  result. 


14  ALGEBRA. 

II.  If  there  are  two  numbers  with  unlike  signs.  Find 
the  difference  of  their  absolute  values,  and  prefix  to  the  result 
the  sign  of  the  greater  number. 

III.  If  there  are  more  than  two  numbers  with  unlike 
signs.  Combine  the  first  two  numbers  and  this  result  with 
the  third  number,  and  so  on;  or,  find  the  sum  of  the 
positive  numbers  and  the  sum  of  the  negative  numbers,  take 
the  difference  between  the  absolute  values  of  these  two  sums, 
and  prefix  to  the  result  the  sign  of  the  greater  sum. 

29.  The  result  in  each  case  is  called  the  sum.  It  is  often 
called  the  algebraic  sum,  to  distinguish  it  from  the  arith- 
metical sum,  that  is,  the  sum  of  the  absolute  values  of  the 
numbers. 

30.  Subtraction  of  Algebraic  Numbers.  In  order  to  sub- 
tract one  algebraic  number  from  another,  we  begin  at  the 
place  in  the  series  which  the  minuend  occupies  and  count,  in 
the  direction  opposite  to  that  indicated  by  the  sign  of  the 
subtrahend,  as  many  units  as  there  are  in  the  absolute  value 
of  the  subtrahend. 

Thus,  the  result  of  subtracting  +  3  from  +  4  is  found  by  counting 
from  +  4  three  units  in  the  negative  direction;  that  is,  in  the  direction 
opposite  to  that  indicated  by  the  sign  +  before  3,  and  is,  therefore,  +  1. 

The  result  of  subtracting  —  3  from  +  4  is  found  by  counting  from 
+  4  three  units  in  the  positive  direction ;  that  is,  in  the  direction 
opposite  to  that  indicated  by  the  sign  —  before  3,  and  is,  therefore,  +  7. 


-5    -4-3-2-1      0+1+2+3+4+5    +6 

The  result  of  subtracting  +  3  from  —  4  is  found  by  counting  from 

—  4  three  units  in  the  negative  direction,  and  is,  therefore,  —  7. 

The  result  of  subtracting  —  3  from  —  4  is  found  by  counting  from 

—  4  three  units  in  the  positive  direction,  and  is,  therefore,  —  1. 


ADDITION    AND    SUBTRACTION.  15 

Collecting  the  results  obtained  in  addition  and  subtrac- 
tion, we  have 

Addition.  Subtraction. 

+4  +  (-3)=+4-3  =  +  l.  +4-(+3)-  +  4-3=+l. 

+4+(+3)-+4+3-+7.  +4-(-3)=+4+3  =  +  7. 

-4  +  (-3)--4-3--7.  -4-(+3)  =  -4-3--7. 

-4  +  (+3)  =  -4+3  =  -l.  ~4-(-3)  =  -4+3  =  -l. 

If  we  employ  the  general  symbols  a  and  h  to  represent 
the  absolute  values  of  any  two  algebraic  numbers,  we  have 

Addition.  Subtraction. 

-\-a-{-{-h)^-\-a-h.  +a-(+5)=-+a-6.  (1) 

+  a  +  (+^)  =  +  «  +  ^.  -\-a~{-h)^-\-a+b.  (2) 

-a^{~h)  =  -a-b.  -a-{+b)==-a-b.  (3) 

-a4-(+5)-:-a  +  6.  -a-(-b)  =  -a+b.  (4) 

From  (1)  and  (3),  it  is  seen  that  subtracting  a  positive 
number  is  equivalent  to  adding  an  equal  negative  number. 

From  (2)  and  (4),  it  is  seen  that  subtracting  a  negative 
number  is  equivalent  to  adding  an  equal  positive  number. 

Hence,  to  subtract  one  algebraic  number  from  another : 

Change  the  sign  of  the  subtrahend,  and  add  the  subtra- 
hend to  the  minuend. 

This  rule  is  consistent  with  the  definition  of  subtraction 
given  in  §  26  ;  for,  if  we  have  to  subtract  —  4  from  +  3,  we 
must  add  -f  4  to  the  subtrahend,  —4,  to  cancel  it,  and  then 
add  +  3  to  obtain  the  minuend  ;  that  is,  we  must  add  +  7 
to  the  subtrahend  to  get  the  minuend,  but  -f  7  is  obtained 
by  changing  the  sign  of  the  subtrahend,  —4,  making  it  +4, 
and  adding  it  to  +  3,  tlie  minuend. 


16  ALGEBRA. 

31.  The  Commutative  Law  of  Addition.  If  we  have  a  group 
of  3  things  and  another  group  of  4  things,  we  shall  have  a 
group  of  7  things,  whether  we  put  the  3  things  with  the 
4  things  or  the  4  things  with  the  3  things. 

That  is,  4  +  3  =  3  +  4. 

If  now  we  have  —  3  to  add  to  +  4,  we  begin  at  +  4  in 
the  series,  count  three  units  to  the  left,  and  arrive  at  +  1  ; 
and  if  we  have  +4  to  add  to  —  3,  we  begin  at  —  3  in  the 
series,  count  four  units  to  the  right,  and  arrive  at  +  1. 

That  is,  +  4  +  (-  3)  =  -  3  +  (+  4). 

Hence,  if  a  and  h  stand  for  any  two  numbers  whatever, 
we  have 

a -\- h  =^  b  -{-  a. 

This  is  called  the  commutative  law  of  addition,  and  may 
be  stated  as  follows  : 

Additions  may  he  performed  in  any  order. 

32.  The  Associative  Law  of  Addition.  If  we  have  several 
numbers  to  be  added,  the  result  will  evidently  be  the  same, 
whether  we  add  the  numbers  in  succession  or  arrange  them 
in  groups  and  add  the  sums  of  these  groups. 

Thus,  a  +  ^>  +  c  +  6?+e 

=  a  +  (^  +  ^)  +  (c?  +  e) 
=  («  +  5)  +  (c  +  cZ+e). 

This  is  called  the  associative  law  of  addition,  and  may 
be  stated  as  follows  : 

The  terms  of  an  expression  m,ay  he  grouped  in  any 
manner. 


ADDITION   AND   SUBTRACTION.  17 

33.  Addition  of  Integral  Expressions.  The  addition  of  two 
integral  expressions  can  be  represented  by  connecting  the 
second  expression  with  the  first  by  the  sign  +•  If  there 
are  no  like  terms  in  the  two  expressions,  the  operation  is 
algebraically  complete  when  the  two  expressions  are  thus 
connected. 

If,  for  example,  it  is  required  to  add  m-\-n—p  to 
a-\-h-{-  c,  the  result  will  be  a -\- h  -\-  c -{- {m -\- n  —p). 

34.  If,  however,  there  are  like  terms  in  the  expressions 
to  be  added,  the  like  terms  can  be  collected;  that  is,  every 
set  of  like  terms  can  be  replaced  by  a  single  term  with  a 
coefficient  equal  to  the  algebraic  sum  of  the  coefficients  of 
the  like  terms. 

If  it  is  required  to  add  Sa'^-f-^a  +  S  to  2a''— 3a  — 4, 
the  result  will  be 

2a2-3a-4  +  (5a^  +  4a  +  3) 

=  2a^  -  3a  -  4  +  Sa'^  -f  4a  +  3  §  32 

-2a^  +  5a='~3a4-4a-4  +  3  §31 

=  (2a''  +  ba')  +  (-  3a  +  4a)  +  (-  4  +  3)         §  32 
=  7a''-{-a-l. 

This  process  is  more  conveniently  represented  by  arrang- 
ing the  terms  in  columns,  so  that  like  terms  shall  stand  in 
the  same  column,  as  follows: 

2a'- 3a  — 4 
5a' +  4a +  3 

7a' +    a-1 

The  coefficient  of  a^  in  the  result  will  be  5  +  2,  or  7  ;  the  coeffi- 
cient of  a  will  be  —  3  +  4,  or  1 ;  and  the  last  term  is  —  4  +  3, 
or-1. 


18  ALGEBRA. 

If  we  are  required  to  find  the  sum  of 

2a?  -  3a^6  +  ^ah'  +  b\  a'  +  ^a'b  -lab'-  2b\ 

-2>o?^a^b-  Sab'  -  W,  and  2a^  +  2a'b  +  6ab'  -  35^ 
we  write  them  in  columns,  as  follows : 

2a'-Sa'b  +  4:ab'+    b' 

a'  +  4:a'b-7ab''-2b' 

-3a^+    a'b~Sab'~W 

2a^  +  2a'b  +  6ab'-3b' 


2a'  +  4:a'b  -8b' 

The  coefficient  of  a^  in  the  result  will  be  2  +  1—3  +  2,  or  +  2 ; 
the  coefficient  of  a'^b  will  be  —  3  +  4  +  1  +  2,  or  +  4  ;  the  coefficient 
of  ab"^  will  be  4  -  7  -  3  +  6,  or  0 ;  and  the  coefficient  of  ¥  will  be 
1  _  2  -  4  -  3,  or  -  8. 

Exercise  3. 
Perform  the  additions  indicated  : 

1.  (+16) +  (-11).  3.    (+68) +  (-79). 

2.  (-15) +  (-25).  4.    (-7) +  (+4). 

5.  (+33) +  (+18). 

6.  (+  378)  +  (+  709)  +  (-  592). 

7.  A  man  has  $5242  and  owes  |2758.     How  much  is  he 

worth  ? 

8.  The  First  Punic  War  began  B.C.  264,  and  lasted  23 

years.     When  did  it  end  ? 

9.  Augustus  Caesar  was  born  B.C.  63,  and  lived  77  years. 

When  did  he  die  ? 

10.  A  man  goes  65  steps  forwards,  then  37  steps  backwards, 
then  again  48  steps  forwards.  How  many  steps  does 
he  take  in  all?  How  many  steps  is  he  from  where 
he  started  ? 


ADDITION    AND    SUBTRACTION.  19 

Exercise  4. 
Perform  the  additions  indicated  : 

1.  {-[-bab)  +  {-bah).  6.    (+7a5)  +  (-5a5). 

2.  (+8m:c)  +  (-2??i.T).         7.    (+ 120  wy) +  (- 95^2/). 

3.  {-l^m.ng)  +  {-1mng).    8.    (- 33a5^)  +  (+lla^>'). 

4.  (_5a;'^)  +  (+8rr^).  9.    {-Ibxy)  +  {-^20xy). 

5.  (+25wy')4-(-18my').  10.    (+ 15aV)  +  (- aV). 

11.  (+5a)  +  (-35)  +  (+4a)  +  (-75). 

12.  (+  4aV)  +  (-  10a:y2)  +  (+  6aV)  +  (-  Qrr^/z) 

-^  {- II a'e) -{-{-{- 20 xyz). 

13.  (+  3a:»  +  (-  4a5)  +  (-  2mn)  +  (+  bx'y) 

Exercise  5. 
Add  the  following  expressions  : 

1.  5a  +  35  +  c,   3a+3^  +  3c,   a  +  35  +  5^. 

2.  7a— 4Z>  +  c,   6a-|-3^-5c,   -12a  +  4c. 

3.  a-|-^  — ^,   ^  +  c  — a,   c  +  a  — 5,   a  +  ^  — ^• 

4.  a-I-25  +  3c,    2a  — 5  — 2c,    b—a  —  c,   c  —  a  —  b. 

5.  a-25-f  3c  +  4a^,    35  -  4c  + 5c?- 2a, 

5c-6c^+3a-45,    7c^- 4a  + 55  -  4c. 

6.  a:'-4a;^  +  5^-3,    2.^^  -  7a;^~  7a:^- 14:r  +  5, 

7.  a;*  — 2a;'+3:c^    a;'  +  a;2_|_^^    4^*^-5a7^ 

2:r'^  +  3a;-4,   -3:r^-2a;-5. 

8.  a'4-3a6^-3a=^5-5V  2a'+ Sa'^^ -6a5'^  -  75^ 

a'-ab''-\-2b\ 

9.  2a5-3aa;'  +  2a'^:r,    12a6  -  6a'a;+ 10a:r', 

ao;^  —  8  a5  —  5  a^x. 


20  ALGEBRA. 

11.  Sx"^ —  xy-{-xz  —  37/'^-\-4:yz  —  z^,  —^x^  —  xy—xz-\-^yz, 

6a;^  —  63/  — 62;,  43/2;  —  5^2  +  3 2^ 
-^x^^y'-\-Zyz-^Zz\ 

12.  m^  —  3  w*w  —  ^rf^r^,   ■\-  mV  +  ^^^^^  —  5m*n, 

7  'w^ri!'  -f  4  mW  —  3  mn*',  —  2  m'^n^  —  3  Tnn^  +  4  re', 
2m?^*  +  2n^4-3m^   —  w^  +  2m^  +  7m*«. 

Exercise  6. 
Perform  the  subtractions  indicated  (§  30)  : 

1.  (+25) -(+16).  3.    (-31) -(+58). 

2.  (-50) -(-25).  4.    (+ 107)  -  (- 93). 

5.  Rome  was  ruled  by  emperors  from  B.C.  30,  to  its  fall, 

A.D.  476.     How  long  did  the  empire  last? 

6.  The  continent  of  Europe  lies  between  36°  and  71°  north 

latitude,  and  between  12°  west  and  63°  east  longi- 
tude (from  Paris).  How  many  degrees  does  it 
extend  in  latitude,  and  how  many  in  longitude  ? 

Exercise  7. 
Perform  the  operations  indicated  : 

1.  (+5a;)-(~4a:).  6.    (+ 17aa;^)  -  (- 24a^3). 

2.  (-3a5)-(+5a5).  7.    (+ 5a'a:)  -  (- 3a=^:r). 

3.  (+3a50-(+10a5'^).        8.    {- /^xy)  -  {-bxy). 

4.  (+15mV)-(-7mV).    9.    (+ 8aa:)  -  (- 3ay). 

5.  (-7^2/) -(-3 ay).  10.    (+ 2«%)  -  (+a^*y). 

11.  (+9a;'')  +  (5a;'^)-(+8a;''). 

12.  (+5 x'y)  -  (-  18a;V)  +  (-  lOo^V). 


ADDITION    AND   SUBTRACTION.  21 

13.  {+llaa^)-{-  ax')  —  (+  24  ax'). 

14.  (-  3  ab)  +  (2  7nx)  -  (—  4  mx). 

15.  (+3a)-(+2^»)-(-4c). 

Exercise  8. 

In   performing   the  following   subtractions  change   the 
signs  of  the  subtrahend  mentally  and  add  : 

1.  From6a  —  2b-ci9ke2a-2b-3c. 

2.  From  3a  — 25  +  3ctake2a  — 7^-c-6. 

3.  From  7:^2 -8a;- 1  take  5a;' -6a; 4- 3. 

4.  From4a;*-3a;'-2a;'-7a;  +  9 

take  a;*  -  2a;'  ~  2a;'^  +  7x  -  9. 

5.  From  2a;'  — 2 aa;4- 3 aHake  a;' —  aa:4-<^'. 

6.  From  a;'  — 3  a;2/  — 2/'  + 3/2  — 22' 

take  x'  +  2xy-\-5xz  —  Sy^  -  2z\ 

7.  Froma'-3a'Z'  +  3a5'-^>' 

take  -d'  +  Sa'b-  Sab'  +  b\ 

8.  From  a;'  — 5  a;y  +  a;z  —  j^'-f  73/2  +  22' 

take  x"^  —  xy  —  xz  -\-  2yz  -{-  S 2'. 

9.  From2aa;'-f3a5a;  — 45'a;+125' 

take  ax'  —  iabx  +  5a;'  —  5  b'x  —  a?. 

10.  From  6 a;'  -  7 a;'y  +  4 a;?/'  -  2y'  —  5a;'  +  a;y  -  4y'  +  2 

take  8  a;'  —  7  a;'2/  -f  a;2/'  -  2/'  +  9  a;'  -  a;y  -f  6  y'  -  4. 

11.  From  a*  -  ^>*  take  ^a'b  -  6a'5'  +  4a5',  and  from  the 

result  take  2  a*  -  4  a'5  +  6  a'b'  +  4  a6'  -  2  5*. 

12.  From  a;y  —  3a;'2/'  +  4a;3/*  —  3/*  take  —  ar^  +  2a;*2/  —  4a;y* 

—  43/^.     Add  the  same  two  expressions,  and  subtract 
their  difference  from  their  sum. 

13.  From  a'5'  -  a^bc  -  Sab^c  -  aV  +  abc'  -  Qb'c' 

take  2  a'bc  -  5  ai'c  +  2  a5c'  —  5  6'c'. 


22  ALGEBRA. 

14.  From  12a  +  Sb-bc  —  2d  take  10a— b -j-4:C —  3d, 

and  show  that  the  result  is  numerically  correct  when 
a  =  6,b  =  4:,c=l,d=5. 

15.  What  number  must  be  added  to  a  to  make  5  ;  and  what 

number  must  be  taken  from  2a^—Qa^b-\-  6ab^—2b^ 
tolesLYea^-ld'b-Sb'? 

16.  From  2  x"^  —  y'^  —  2 xy -\- z^  take  x"^  —  y"^  -}-2xy  —  2^ 

17.  From  12ac  +  8cd-9  take  -7ac  —  9cd-\-8. 

18.  From— Qd'-\-2ab  —  Sc'' take  4:a'  +  eab-Ac\ 

19.  From  9x1/  —  4ri;  —  3y  +  7  take  Sxy  —  2x -{-3y -j-6. 

20.  From  —  d^bc  —  ab^c  +  abc^  —  abc 

take  a^bc  +  ctb'^o  —  aZ>c^  +  a5d7. 

21.  From  7.T'-2a:  +  4take  2:r'^  +  3a;-l. 

22 .  From  Za?  -\-2xy  —  y"^  take  —  a;''  —  3 a:y  +  3 3/^  and  from 

the  remainder  take  8  a;'  +  4a;y  —  5yl 

23.  From  ax^  —  by"^  take  cx^  —  dy'^. 

24.  From  ax  ■\- bx -\- by  -\-  cy  take  ax  —  bx  —  by  -\-  cy. 

25.  From  5a;'^  + 4a: -4y4-3y' take  5a:2_  3^,  _|_  3^_|_^2^ 

26.  From  a^b"^  +  12  aSc  —  9  ax^  take  4 ai'^  —  6  acx  -j-  3  aV. 

27.  From  a'  -  2a5  +  c'^  -  36'  take  2a'  -  2a6  +  35'. 

28.  From  the  sum  of  the  first  four  of  the  following  expres- 

sions, a' +  6' +  c' +  c?',  d''-^b^-\-c\  a^-&-\-b''-d}, 
a^  —  b''-\-c^^d\  5'  +  c'  +  c?'-a',  take  the  sum  of 
the  last  four. 

29.  From  2a;'  -  2y''  -  2'  take  3y'  +  2a:'  -  z^,  and  from  the 

remainder  take  3  z'  —  22/'  —  a:'. 

30 .  From  a'  —  2  a'c  +  3  a(?  take  the  sum  of  a'c  —  2  a^  +  2  ac' 

and  a'  —  ac'  —  d^c. 


ADDITION    AND    SUBTRACTION.  23 

35.  Kules  for  removing  Parentheses.  From  (§  30),  it  ap- 
pears that 

a  -f  (+  6)  =  a  +  6.  a  —  {:\-b)  =  a-b. 

a-{-{—b)  =  a  —  b.  a  — (— b)  =  a-{- b. 

The  same  rules  for  removing  parentheses  hold  true 
whether  one  or  more  terms  are  inclosed.  Hence,  when  an 
expression  within  a  parenthesis  is  preceded  by  a  plus  sign, 
the  parenthesis  may  be  removed. 

When  an  expression  within  a  parenthesis  is  preceded  by 
a  minus  sign,  the  parenthesis  may  be  removed  if  the  sign  of 
every  term  within  the  parenthesis  is  changed. 

Thus,  a  -{-  (b  —  c)  =  a  -]-  b  —  c. 

a  ~  (b  ~  c)  =  a  ~  b  -{-  c. 

36.  Expressions  may  occur  with  more  than  one  paren- 
thesis. In  such  cases  parentheses  of  different  shapes  are 
used,  and  the  beginner  when  he  meets  with  a  (  or  a  [  or 
a  {  must  look  carefully  for  the  other  part,  whatever  may 
intervene ;  and  all  that  is  included  between  the  two  parts 
of  each  parenthesis  must  be  treated  as  the  sign  before  it 
directs,  without  regard  to  other  parentheses.  It  is  best  to 
remove  each  parenthesis  in  succession,  beginning  with  the 
innermost  one.     Thus, 

(1)  5a-  J  -  3  a  —  [3  a  ~  (2  a  —  a  -  5)  -  a]  +  a  J 
=  5(2-  j-3a  — [3a  -  (2a -a  + ^)  -  a]  +  a? 
=  5a—  \—  2> a  —  \?> a -—  2a -{•  a  —  b  —  a^  -\-  a\ 

—  ^a~  \~2>a  —  2>a-\-2a  —  a-\-b-]-a-\-a\ 
=  5a  +  3a  +  3a— 2a  +  a  —  6  — a  —  a 

—  ba-{-?>a-\-'^a  —  2a-{-a  —  a  —  a—b 
^Sa-b. 

Note.  The  sign  —  which  is  written  in  the  above  problem  before 
the  first  term  a  under  the  vinculum  is  really  the  sign  of  the  vinculum, 
—  a  —  b  meaning  the  same  as  —  (a  —  b). 


24  ALGEBRA. 


Exercise  9. 


Simplify   the   following   expressions    by   removing    the 
parentheses  and  collecting  like  terms : 

1.  {^a-\-b)-\-{b-^c)-{a  +  c). 

2.  i2a  —  h-c)  —  {a-2b-\-c). 

3.  (2x-y)-i2y-z)-{2z~x). 

4.  {a~x-y)  —  {h  —  x-]-y)^{c  +  2y). 

5.  {2x  -  y  -^^z)  -^  {-  X  -  y  -  ^z)  -  {2>x  -2y  -  z). 

6.  (3a-5+7c)-(2a  +  3^)-(55-4c)  +  (3^-a). 

7.  l-(l-a)  +  (l-a  +  a^)-(l-a  +  a^-a'). 

8.  a-  {2h-{^c  +  2b)-a\, 

9.  2a-  {b~{a-2h)\. 

10.  3a-  J^>  +  (2a-6)  — (a-5)|. 

11.  7a-[3a- J4a-(5a-2a)J]. 

12.  2a;  +  (y~32)-K3a:-2y)  +  2J+6a;-(4y-32). 

13.  |(3a  -  26)  +  (4c  -  a)\-\a- {2b -?>a)  -  c\ 

■^\a-{b-bc-a)\. 

14.  a-[2a-f  (3a-4a)]-5a-J6a-[(7a  +  8a)-9a]S. 

15.  2a-(35  +  2c)-[5^>-(6c-66)  +  6c 

-J2«-(c  +  2^»)n. 

16.  a-[25  +  J3c-3a-(a  +  ^)J  +  f2a-(^>-f-c)|]. 

17.  16 -a; -[7a; -58:?; -(9  a; -3a; -6a;)  J]. 


18.  2a-[36+(25-c)-4c  +  J2a-(35-c-25)|]. 

19.  a-[26  +  f3c-3a-(a  +  6)S4-2a-(5  +  3c)]. 

20.  a-[55-{a-(3c-3^')  +  2c-(a-26-c)j]. 


ADDITION    AND    SUBTRACTION.  25 

37.  Eules  for  Introducing  Parentheses.  The  rules  for  in- 
troducing parentheses  follow  directly  from  the  rules  for 
removing  them  : 

1.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  plus  placed  before  it. 

2.  Any  number  of  terms  of  an  expression  may  be  put 
within  a  parenthesis,  and  the  sign  minus  placed  before  the 
parenthesis ;  provided  the  sign  of  every  term  within  the 
parenthesis  is  changed. 

It  is  usual  to  prefix  to  the  parenthesis  the  sign  of  the 
first  term  that  is  to  be  inclosed  within  it. 

Exercise  10. 

Express  in  binomials,  and  also  in  trinomials : 

1.  2«-3^>-4c  +  c?+3e-2/ 

2.  a-2x^^y-Zz-2h^c. 

3.  aH3a*-2a'-4a'  +  a-l. 

4.  -2>a-2h-^2c-bd—e-2f. 

5.  ax  —  hy  —  cz  —  bx  -{-  cy  -{-  az. 

6.  2x''  -  3a;V  +  ^^Y  -  ^ ^V  +  ^V'  -  '^f- 

7.  Express  each  of  the  above  in   trinomials,   each   tri- 

nomial having  its  last  two  terms  inclosed  by  inner 
parentheses. 

Collect  in  parentheses  the  coefficients  of  x,  y,  z  in 

8.  2ax  —  ^ay-\-4ibz  —  ^hx  —  2cx  —  ^cy. 

9.  ax  —  hx-{-2ay-\-^y-\-^az  —  2>hz  —  2z. 

10.  ax  ~  2hy -{- b cz  —  4:hx  —  Scy-{-az  —  2cx  —  ay-\-^hz. 

11.  \2ax  +I2ay  -\-  ^hy  -I2hz  -lb  ex  -\-  Qcy  +  2>cz. 

12.  2ax  —  ^by  —  7 cz  —  2bx  +  2cx  -\-  Scz  —  2cx  —  cy  —  cz. 


CHAPTER  III 

MULTIPLICATION.  vv^cXa^  ^  . 

Integkal  Expressions. 

38.  Definition  of  Multiplication.  The  process  of  finding 
the  result  when  a  given  number  is  taken  as  many  times  as 
there  are  units  in  another  number  is  called  multiplication, 
and  the  result  is  called  the  product. 

This  definition  fails  when  the  multiplier  is  a  fraction,  for 
we  cannot  take  the  multiplicand  a  fraction  of  a  time.  We 
therefore  consider  what  extension  of  the  meaning  of  multi- 
plication can  be  made  so  as  to  cover  the  case  in  question. 
When  we  multiply  by  a  fraction,  we  divide  the  multiplicand 
into  as  many  equal  parts  as  there  are  units  in  the  denomi- 
nator and  take  as  many  of  these  parts  as  there  are  units 
in  the  numerator.  If,  for  instance,  we  multiply  8  by  f ,  we 
divide  8  into  four  equal  parts  and  take  three  of  these  parts, 
getting  6  for  the  product.  We  see  that  f  is  I  of  1,  and  6 
is  f  of  8 ;  that  is,  the  product  6  is  obtained  from  the  mul- 
tiplicand 8  precisely  as  the  multiplier  f  is  obtained  from  1. 

Again,  in  6  X  8  =  48, 

the  multiplier  6  is  1  +  1  +  1  +  1  +  1  +  1, 

and  the  product       48  is  8  +  8  +  8  +  8  +  8  +  8. 

Hence  we  have  for  the  general  definition  of  multipli- 
cation, 

The  operation  of  finding  from  two  given  numbers,  called 
multiplicand  and  multiplier,  a  third  number  coWq^l  product, 


MULTIPLICATION.  27 

which  is  formed  from  the  TnuUiplicand  as  the  multiplier  is 
formed  from  unity. 

39.  Law  of  Signs  in  Multiplication.  By  the  definition  of 
multiplication, 

since  +  3  -  (-M)  +  (+  1)  +  (+  1), 

.-.  3x(+4)  =  (+4)4-(+4)  +  (-f4) 
=  +  12, 
and  3  X  (-4)  =  (-4) +  (-4) +  (-4) 

=  -12. 
Again,  since         -  3  =  (-  1)  +  (-  1)  +  (-  1), 
,.  (_3)x4  =  (-4)  +  (-4)4-(-4) 
=  -12, 
and  (-  3)  X  (-  4)  =  -  (-  4)  -  (-  4)  -  (-  4) 

=+4+4+4 
=  +  12. 

If  we  use  a  to  represent  the  absolute  value  of  any  num- 
ber, and  b  to  represent  the  absolute  value  of  any  other 
number,  we  shall  have 

(+a)x(+5)  =  +  a^>.  (1) 

(+  a)  X  (-  5)  =  -  ah,  (2) 

(-a)x(+^')=-a5.  (3) 

(-  a)  X  (-  6)  =  +  ah.  (4) 

Hence,  in  finding  the  product  of  two  algebraic  numbers, 
Like  signs  give  +,  and  unlike  signs  give  —. 

40.  It  will  be  seen  from  the  above  cases  that  the  absolute 
value  of  the  product  is  independent  of  the  signs  of  the  factors, 
and  that  the  sign  of  the  product  is  independent  of  the  order 
of  the  factors. 


28  ALGEBRA. 

41.  The  Oommutative  Law  of  Multiplication.  If  we  have 
five  lines  of  dots  with  ten  dots  in  a  line,  the  whole  number 
of  dots  will  be  expressed  by  5  X  10. 


If  we  consider  the  dots  as  ten  columns  with  five  dots  in 
a  column,  the  number  will  be  expressed  by  10  X  5. 

That  is,  5  X  10  =  10  X  5. 

Hence,  if  a  and  h  stand  for  any  two  positive  numbers, 

ah  =  ha. 

42.  The  Distributive  Law  of  Multiplication.  The  expression 
4  X  (5  -f  3)  means  that  we  are  to  take  the  sum  of  the  num- 
bers 5  and  3  four  times.  The  process  can  be  represented 
by  placing  five  dots  in  a  line,  and  a  little  to  the  right  three 
more  dots  in  the  same  line,  and  then  placing  a  second, 
third,  and  fourth  line  of  dots  underneath  the  first  line  and 
exactly  similar  to  it. 

•  ••••  ••• 

•  ••••  ••• 

•  ••••  ••• 

•  ••••  ••• 

There  are  (5  +  3)  dots  in  each  line,  and  4  lines.  The 
total  number  of  dots,  therefore,  is  4  X  (5  +  3). 

In  the  left-hand  group  there  are  4x5  dots,  and  in  the 
right-hand  group  4x3  dots.  The  sum  of  these  two  num- 
bers (4  X  5)  +  (4  X  3)  must  be  equal  to  the  total  number  ; 
that  is, 

4  X  (5  +  3)  =  (4  X  5)  +  (4  X  3).  (1) 


MULTIPLICATION.  29 

Again,  the  expression  4  X  (8  —  3)  means  that  3  is  to  be 
taken  from  8,  and  the  remainder  to  be  multiplied  by  4. 
The  process  can  be  represented  by  placing  eight  dots  in  a 
line  and  crossing  the  last  three,  and  then  placing  a  second, 
third,  and  fourth  line  of  dots  underneath  the  first  line  and 
exactly  similar  to  it. 

^^4 

><»<»» 

*^* 

*^^ 

The  whole  number  of  dots  not  crossed  in  each  line  is 
evidently  (8  —  3),  and  the  whole  number  of  lines  is  4. 
Therefore  the  total  number  of  dots  not  crossed  is 

4x(8-3). 

The  total  number  of  dots  (crossed  ar  1  not  crossed)  is 
(4x8),  and  the  total  number  of  dots  crossed  is  (4x3). 
Therefore  the  total  number  of  dots  not  crossed  is 

(4  X  8)  -  (4  X  3). 
Hence,         4  X  (8  -  3)  =  (4  X  8)  -  (4  X  3).  (2) 

If  a,  b,  and  c  stand  for  any  positive  numbers,  (1)  becomes 

a  X  (b -i- c)  =  ab -{-  ac, 
and  (2)  becomes 

a  X  (b  —  c)  —  ab  —  ac. 

This  is  the  distributive  law  of  multiplication  and  may  be 
stated  as  follows : 

In  multiplying  a  compound  expression  by  a  simple  ex- 
pression, the  result  is  obtained  by  multiplying  each  term  of 
the  compound  expression  by  the  simple  expression,  and  writ- 
ing down  the  successive  products  with  the  same  signs  as 
those  of  the  original  terms. 


30  ALGEBRA. 

43.  The  Associative  Law  of  Multiplication,  The  product 
of  three  or  more  factors  is  evidently  the  same  in  whatever 
way  the  factors  are  grouped.     Thus, 

7  X  (3  X  5)  =  3  X  (5  X  7)  =  5  X  (7  X  3)  =  105. 

Hence,    cX{aXh)  =  aX  {h  x  c)  =  hx  {c  x  a)-=  ahc, 
where  a,  6,  and  c  stand  for  any  positive  numbers. 

This  is  called  the  associative  law  of  multiplication,  and 
may  be  stated  as  follows : 

The  factors  of  a  product  may  he  grouped  in  any  manner. 

44.  Since  (§  40)  the  absolute  value  of  a  product  is  inde- 
pendent of  the  signs  of  its  factors,  and  the  sign  of  a  product 
is  independent  of  the  order  of  its  factors,  it  is  evident 
that  the  commutative,  the  distributive,  and  the  associative 
laws  of  multiplication  apply  to  algebraic  as  well  as  to 
arithmetical  numbers. 

45.  The  Index  Law  of  Multiplication. 

Since  g^  =  aa,  and  a^  =  aaa,  §  19 

a^X  a^  =  aaX  aaa  =  a*  =  a^+' ; 
a*  X  a  =  aaaa  X  a  =  a^  =  a*+\ 
If  a  stands  for  any  number,  and  m  and  n  for  any  integers, 

since       a**  =  aaa to  m  factors, 

and         a**  =  aaa to  n  factors, 

a""  X  a*"  =  (aaa to  m  factors)  X  (aaa ton  factors) 

=  aaa to  (m  +  w)  factors, 

=  a'"+".     Hence, 

The  index  of  the  product  of  two  powers  of  the  same  num- 
ber is  equal  to  the  sum  of  the  indices  of  the  factors. 

46.  The  commutative,  the  distributive,  the  associative, 
the  index  laws,  and  the  law  of  signs,  constitute  the  funda- 
mental laws  of  Algebra. 


MULTIPLICATION.  31 

47.  Multiplication  of  Monomials.  When  the  factors  are 
single  letters,  the  product  is  represented  by  simply  writing 
the  letters  without  any  sign  between  them. 

Thus,  the  product  of  a,  h,  and  c  is  expressed  by  ahc ;  and 
the  product  of  4  a,  56,  and  3  c  is 

4aX55x  3c  =  4x  bx2>ahc  =  mahc.  §41 

Note.  We  cannot  write  453  for  4  X  5  x  3,  because  another  mean- 
ing has  been  assigned  in  Arithmetic  to  453 ;  namely,  400  +  50  +  3. 
Hence  between  arithmetical  factors  the  sign  X  must  be  written. 

48.  By  the  law  of  signs,  we  have 

{-a)x{-h)^-{-ah, 
and  (-j-  ah)  X  (— c)  =  —  ahc  ; 

that  is,         (—  a)  X  (— h)  X  (— c)  =  —  ahc. 

Hence,  the  product  of  an  even  number  of  negative  factors 
will  be  positive,  and  the  product  of  an  odd  number  of  nega- 
tive factors  will  be  negative. 

49.  The  product  of  a'^h  and  a^h"^  is 

a'h  X  a'h'  =  aWhh'  =  a'+'h'+'  =  a'h\ 

50.  To  multiply  one  monomial  by  another,  therefore, 

Mnd  the  product  of  the  coefficients,  and  to  this  product 
annex  the  letters,  giving  to  each  letter  in  the  product  an  index 
equal  to  the  sum  of  it^  indices  in  the  factors. 

51.  (ahcf  means  ahc  X  ahc,  which  equals  aahhcc,  or  a^h^c^. 
In  like  manner,  (ahcY  =  a^'h^'c*'.     That  is, 

The  nth  power  of  the  product  of  several  factors  is  equal  to 
the  product  of  the  nth  powers  of  the  factor's. 


32  ALGEBRA. 

52.   Polynomials  by  Monomials.     We  have  (§  42), 

a(b  -]-  c)  =  ab  -\-  ac ; 

and,  a(b  —  c-^d—e)  =  ab  —  ac-\-ad—ae. 

To  multiply  a  polynomial  by  a  monomial,  therefore, 

Multiple/  each  term  of  the  polynomial  by  the  monomial, 
and  add  the  partial  products. 

». 

Exercise  11. 
Find  the  product  of 

1.  -17  and  8.  4.    -  18  and  -  5. 

2.  -  12.8  and  25.  5.   43  and  -  6. 

3.  3.29  and  5.49.  6.   457  and  100. 

7.  (-358 -417)  and -79. 

8.  (7.512- {-2.894i)  and  (-6.037  + J  13.963 i). 

9.  13,  8,  and  -  7. 
le.'  -38,  9,  and  -6. 

11.  -20.9,-1.1,  and  8. 

12.  -78.3, -0.57, +1.38,  and -27.9. 

13.  -  2.906,  -  2.076,  -  1.49,  and  0.89. 

Exercise  12. 
Find  the  product  of 

lo   6«and— 2a.  3.    3  arr  and  —  4  Jy. 

2.    5mn  and  9m.  4.   —8  cm  and  dn. 


MULTIPLICATION.  33 

5.  ~1  ah  and  2ac.  8.    3aV  and  7aV. 

6.  bm^x  and  2>mx^.  9.    7a,  —45,  and  —8c. 

7.  5 a*"  and  -  2a^  10.    8a5^  3ac,  and  -  ^c\ 

11.  21  ah,—  39  mp,  and  18  op. 

12.  6a6y,  2hy,  and  -  Sa'^y.  » 

13.  7m'lr,  3  wa;'^,  and  —  2mq. 

14.  -  3^^^  6j9Y  and  8pY- 

15.  2a'mV,  3amV,  and  4a'wa:'*. 

16.  60:^2',  —  9^y2^  and  3a:V- 

17.  ^ax,  2  am,  —Amx,  and  61 

18.  7am^  35V,  —  4a5,  a^hn,  —  2hH,  and  —  mn^. 

19.  2a5^  -5a'5,  -3a5,  and  7a. 

Exercise  13. 
Find  the  product  of 

1.  (4a' -36)  and3a6. 

2.  (8«'-9a6)and  3al 

3.  (3a;'  -  ^y"  +  bz")  and  2x'y. 

4 .  (a'a;  —  5  a^x^  +  aa:*  +  2  re*)  and  ax'^y. 

5.  (- 9a^  +  3a'5'  - 4a'5^  -  h')  and  -  3a5*. 

6.  (3rc'  — 2a;'?/-7a;3/'  +  2/')  and-5a:y 

7.  (-  4a:2/'  -f  5 x'y  +  8a;')  and  -  3  x^y. 

8.  (-3  +  2a6  +  a'6')  and-a*. 

9.  (-  z  —  2a;2'  +  5a;V  —  Ga^^y'  +  3a:yz)  and  -  3a;V- 


34  ALGEBRA. 

63.  Polynomials  hj  Polynomials.  If  we  have  m-{n-{'p 
to  be  multiplied  by  a  +  ^  +  ^,  we  may  substitute  Mior  the 
multiplicand  m-\-n-\-p.     Then 

(a  +  5  +  c)  Jf  =  a  Jf  +  bM-{-  cM.  §  42 

If  now  we  substitute  for  J[f  its  value  m'{-n  -\-p^  we  shall 
have 

a{m-\-n-[-p)-\-h{m  +  n-{-p)-]-c(m-{-n-{-p) 
=  am  +  a?2  +  «p  +  ^^  -\- bn -\- bp  -\-  cm -{-  en -\-  cp. 

That  is,  to  find  the  product  of  two  polynomials, 

Multiply  every  term  of  the  multiplicand  by  each  term  of 
the  multiplier,  and  add  the  partial  products. 


54.  In  multiplying  polynomials,  it  is  a  convenient  ar- 
rangement to  write  the  multiplier  under  the  multiplicand, 
and  place  like  terms  of  the  partial  products  in  columns. 

(1) 


5a  - 
3a  - 

-  6   b 

-  4:    b 

15  a^- 

-I8ab 
-20ab  +  2U^ 

15a'- 

-S8ab  +  2ib' 

We  multiply  5  a,  the  first  term  of  the  multiplicand,  by  3  a,  the 
first  term  of  the  multiplier,  and  obtain  15  a^ ;  then  —  6  &,  the  second 
term  of  the  multiplicand,  by  3  a,  and  obtain  —  18  a6.  The  first  Hne 
of  partial  products  is  15  a^  —  18  ab.  In  multiplying  by  —  4  6,  we 
obtain  for  a  second  line  of  partial  products  —  20aZ>  +  24  6^  which  is 
put  one  place  to  the  right,  so  that  the  like  terms  —  18  a6  and  —  20  a6 
may  stand  in  the  same  column.  We  then  add  the  coefficients  of  the 
Uke  terms,  and  obtain  the  complete  product  in  its  simplest  form. 


MULTIPLICATION.  35 

(2)  Multiply  4:X-\-S  +  5x'  —  6x'  by  4  -  6a:'  -5x. 

Arrange  both  multiplicand  and  multiplier  according  to 
the  ascending  powers  of  x. 

3+    4x-\-    5a;'-    6.r' 
4-    5a:-    6a:' 

12  +  16a:  +  20a;'- 24 a;^ 

-  15a:  -  20a:'  -  25a:^  +  30a:* 

-  18a;'  -  24a:'  -  30a:*  +  36a;^ 

12+       a; -18a:' -73a;'  +  36a:^ 

(3)  Multiply  1  +  2a;  +  a:*  -  3a;'  by  a:'  -  2  -  2a:. 

Arrange  according  to  the  descending  powers  of  x. 

a:*-3a;'  +  2a:  +1 
x'-2x  -2 


a;^-3a:^  +  2a;*+    a:' 

-2x^  +  6a;' -4a:' -2a: 

-2a:*  +6a:'-4a:-2 

x'  -  5a;^  +  7a:'  +  2a:'  -  6a:  -  2 

(4)  Multiply  a'  +  5'  +  c'  —  a5  —  6c  —  ac  by  a  -j-  5  -f  e. 

Arrange  according  to  descending  powers  of  a. 

a'  — a  6  — a  c  4-   ^^—      ^^+   ^^ 
a  -\-     b-\-     c 

a?  —  a'5  —  a^c  -\-  ah^  —    abc  +  ac' 

+  a'6  -ah'—    ale  -\-b^ -hc-^-bc^ 

+  a'c  —    abc  —  ac^  +  he  —  be'  +  c' 

a'  -^abc  +6'  +c' 

NotHo  The  student  should  observe  that,  with  a  view  to  bringing 
like  terms  of  the  partial  products  in  columns,  the  terms  of  the  multi- 
plicand and  multiplier  are  arranged  in  the  same  order. 


36  ALGEBRA. 

Exercise  14. 

Multiply : 

1.  rr^-4by  :r'  +  5.  3.    a' -\- a^x"  +  x' hj  o?  -  x\ 

2.  y  — 6  by  y  4- 13.  4.    x^ -\- xy -]- y^  hj  x  —  y . 

5.  2x  —  y  hj  X  -\-  2y. 

6.  2a;'  +  4:r*^  +  8a:  +  16by  3a;-6„ 

7.    x^-\-x^-\-x  —  lhy  x  —  1.     8.    o;^  —  3a:r  by  a:  + 3  a. 
9.    25'^  +  3a5  — a''by-5^.  +  7a. 

10.  2a  +  Z>bya  +  2&.  12.    a' -a^>  +  5' by  a  + 6. 

11.  a^-\-ab-\-b''hY  a-h.     13.    2a5  -  5^*^  by  3a^  -  4a5. 

14.  -a3  +  2a^5-5'by  4a^  +  8a5. 

15.  a'  +  ah-\-h''hj  a^-ah-\-h\ 

16.  a^-3a'^^>  +  3a5^-^^by  a'-2a5  +  ^>l 

17.  x-{-2y  -^zhj  x-2y-{-^z. 

18.  2a;'  +  3a:y  +  43/'^by  3:t2-4rry  +  3/z. 

19.  a;^ -f  ^y  +  y'^  by  a;^  +  ^2  +  2^- 

20.'    a}  +  h''-\-c^~ah  —  ac  —  hchja-\-h-\-c. 

21.  :&'■*  —  a;3/  +  2/^  +  ^  +  y  +  1  by  57  +  y  —  1. 

Arrange  the  multiplicand  and  multiplier   according  to 
the  descending  powers  of  a  common  letter,  and  multiply : 

22.  5.r  +  4a;^  +  r?;^  — 24by  a;'+ll-4a:. 

23.  x^-^llx  —  4:X^  —  2^hjx''-\-b-\-^x. 

24.  a;*  +  a;^-4a;-ll  +  2:r' bya;^-2a;  +  3. 

25.  -bx'-x''-x-\-:^-[-nx^hjx''~2-2x, 

26.  3a:  +  a;'-2a;'-4by  2a:  +  4^  +  3rr^  +  l. 


MULTIPLICATION.  37 

27.  oa*  +  2a'b'  +  ab'  -3a'b  by  ba'b  -  2ab^i-Sa'b' i-b*. 

28.  4aV-32a2/*-8ay  +  16ay  by  ay  +  4ay  +  4ay. 

29.  3m'  +  3 w'  +  9mw'  +  Qm'w  by  6mV  -  2mn* 

—  6mW  +  2m*n. 

30.  6a^^  +  Sa'b'  -  2ab'  -{-b' hj  ia' -2ab' ~S  b\ 

Find  the  product  of : 

31.  x  —  S,x—l,x-\-l,&ndx-\-S. 

32.  x''-x  +  l,x^  +  x+l,SLndx^  —  x''-}-l. 

33.  a'  +  abi-  b\  a?  -  ab -^  b\  and  a*  -  a'^^'^  +  b\ 

34.  4a^ - 4a2^>  +  ab\  4a'  +  3a^>  +  ^'^  and  2a^b  +  61 

35.  a; +  «,  ^  + 2a,  a:  — 3a,  a:  — 4a,  and  a:  4- 5a. 

36.  9a''  +  5^27a'-5^27aH^^  and81a*~9a'Z>'  +  ^>*. 

37.  From  the  product  of  y"^  —  2yz  —  ^  and  y^  +  2yz  —  2' 

take  the  product  of  y^  —  yz  —  2z^  and  y'^-{-yz  —  2z^. 

38.  Find  the  dividend  when  the  divisor  =  3  a'  —  a5  —  3  5^ 

the  quotient  =  a^b  —  26^  the  remainder  —  —  2ab^ 
-6b\ 

The  multiplication  of  polynomials  may  be  indicated  by 
inclosing  each  in  a  parenthesis  and  writing  them  one  after 
the  other.*  When  the  operations  indicated  are  actually  per- 
formed, the  expression  is  said  to  be  simplified. 

Simplify : 

39.  (a  +  b-  cXa  +  c-  b)(b  -\-c-  a){a  -^b-\-c). 

40.  (a  +  b){b  +  c)  -  (c  +  d){d  ■^d)-{a-\-  c)(b  -  d). 

41.  (^a-\-b-\-c-\-dY-\-{a-b  —  c-\-dy 

■\-{a  —  b-\-c-dy-{-{a-\-b-c-  d)\ 


38 


ALGEBRA. 


42.  {a-\-h  +  cy~a(h-^c~a)-b{a^c-b)-c{a-{-b-c). 

43.  {a-h)x~{b-\-c)a-\(h-x){h-a)~{h-c){b-{-c)\. 

44.  (mi-n)m~l(m~ny-(n~m)n]. 

45.  (a-b  +  cy~{a(c~a~b)~[b(a  +  b  +  c) 

~c(a  —  b  —  c)]l. 

46.  (p'  +  q')r~(p  +  qXplr-q}~qlr~p]). 

47.  (9:i;y  -  4:y'Xx'  -  f)  ~\2>xy-  2f\  \d>x{x''-{-  f) 

~2y{f-\-2>xy~x')\y. 

48.  dj'  ~\2ab  -[-(^a-^lb  -  c]){a-~\b  ~  c])-\-2ab] 

-Uc\-{b  +  c)\ 

49.  \ac-{a-b)(b^c)\-b\b--{a-c)\. 

50.  ^{a-b)x~cy]~2\a{x-y)~bx] 

~  {^ax~(bc  —  2a)y]. 

51.  (:^-  l)(a;-2)  -  ^x{x-\-^)  +  2  J(:r  +  2)(a;+  1)  -  3}. 

52.  \{2a  +  by  +  {a-2by\x\(2,a-2by-{2a-Uy\. 

53.  4(a-35)(a  +  3^)-2(a-63)2-2(a'^  +  65^). 

54.  a;X^^  +  yy  ~  2xhjXx  +  y){x  -y)-  (x'  -  yj. 

55.  16(a^  +  5^)(a^  -  b'^)  -(2a-  3)(2a  +  3)(4a^  +  9) 

+  (2  5-3)(25  +  3)(4^»^  +  9). 


CHAPTER  IV. 
DIVISION. 

Integral  Expressions. 

55.  Definition  of  Division.  In  division  the  product  and 
one  factor  are  given,  and  the  other  factor  is  required.  We 
may  therefore  take  for  the  general  definition  of  division 

The  operation  by  which  when  the  product  and  one  factor 
are  given  the  other  factor  is  found. 

With  reference  to  this  operation  the  product  is  called 
the  dividend,  the  given  factor  the  divisor,  and  the  required 
factor  the  quotient. 

56.  Law  of  Signs  for  Division. 

Since  (+  a)  X  (+  5)  =  +  ah,     :.  +  a5  ^  (-f  a)  =  +  5. 

Since  (+  a)  X  (—b)  =  —  ab,     .'.  —  ah -^  {-\- a)  =  —  b. 

Since  (—  a)  X  (+  ^)  =  —  Cih,     :.  —ah-^(—a)  =  -\-h. 

Since  (—  a)  X  (~h)  =  -\-  ah,     .'.  -\-  ah  -^  {—  a)  ~  —  b. 

That  is,  if  the  dividend  and  divisor  have  like  signs,  the 
quotient  has  the  sign  + ;  and  if  they  have  unlike  signs, 
the  quotient  has  the  sign  — .     Hence,  in  division. 

Like  signs  give  + ;  unlike  signs  give  — . 

57.  Index  Law  for  Division.  If  we  have  to  divide  a^  by 
a^,  a^  by  d^,  a^  by  a,  a^  by  a^,  we  write  them  as  follows : 

a^         aa 

a^      aaaaaa 


aaaa 


40 


ALGEBRA. 

^= 

aaaa 

=  aaa  = 

a'  = 

a'-'; 

a 

a 

a'_ 

aa 

_  1  _ 

1  _ 

1 

a' 

aaaaa      aaa 

a' 

a'-' 

68.   To  divide  one  monomial  by  another,  therefore, 
Write  the  dividend  over  the  divisor  with  a  line  between 

them :  if  the  expressions  have  common  factors,  remove  the 

common  factors . 

Thus,    —Ji  =  2xy',  ——  =  'Jab;  ^      ^-^x^yz; 

2x  2a  —Qx^yh 

sa'--'      '  Sx^-y-'         '^- 

In  the  last  example,  (p —  i) —  {p —  6)=p —i—p  +  6  =  2  a,nd 
(r  +  3)  -  (r  -  1)  =  r  +  3  -  r  +  1  =  4. 

Note.  Since  a»»  -^  a»*  =  1,  and  by  the  rule  =  a**-"  =  a",  it  follows 
that  a^  =  1.  Hence,  any  letter  in  the  quotient  with  zero  for  an  index 
may  be  omitted  without  affecting  the  quotient. 

59.  To  divide  a  polynomial  by  a  monomial,  we  have,  by  the 
distributive  law,  the  following  rule  : 

Divide  each  term  of  the  dividend  by  the  divisor,  and  add 
the  partial  quotients. 

mi         ^ab  -\-  A:ac  —  ^ad  _  8 a^  ,  4 ac      %ad 
^^'  2a  ~2^      ~2^~^ 

=  4:b-i-2c-Sd. 
9 a'b'x -  12 a^bx'  -Sa'x_9 a*b^x     12 a'bx'     3 d'x 
Sa'x  Sd'x        Sa'x       Sa'x 

=  Sa'b'-4cabx-l. 


2  a;'"-'  2  a;'"-'      2:^2"- 


3^2«+2_2^«+l_ 


Note.  Here  we  have  4n  +  l  — (2n  — l)  =  4n  +  l  — 2n  +  l  =  2n4-2, 
and  3n  —  (2n  —  l)  =  3n  —  2n  +  l==n  +  l,  as  indices  of  a;  in  the  first 
and  last  terms  of  the  quotient  respectively. 


DIVISION. 


41 


Exercise  15. 
Perform  the  operations  indicated  : 


1. 


2. 


3. 


4. 


+  264 
+  4  ■ 

-3840 


+  3840 
+  30  ' 

-2568 
+  12  ' 


5. 


6. 


7. 


8. 


106.33 
-4.9" 

-42.435 

+  34.5  ' 

-264 


+  24 
-3670 


85 


9. 


10. 


11. 


12. 


+  6.8503 
-61 

-  7.1560 

+  324  * 

-1 
-3.14159 

-0.31831 
-31.4159' 


1. 


Perform 
-\-ah 


Exercise  16. 

the  operations  indicated  : 
10  ah 


4. 


5. 


6. 


+  « 
-{-ab 
—  a 

—  ah 
+  a" 

—  ah 

—  a 

6m.y 

12  a* 

-3a 

19. 
20. 


7. 


9. 


10. 


11, 


12. 


4a''^mV 
5  a^m^x 

42rVV 
7a;^V' 


2hc 

x' 
-x' 

—  12  am 

—  2m.  ' 

S5abc 
bhd  ' 

ahx 
bahy 

21  a' 
~2>a;' 


21. 


22. 


—  3  hmx 

13. 

4:ax^ 

ah'(^ 

14. 

abc 

15. 

m^/>V 

mp^x^ 

16. 

-  51 abdy^ 

Udy 

17. 

22b  m^y 

2b  mf 

18. 

30  a;  V 

-5r^y 

-3a'^»' 

'c'd^^ 

-a'b'cd' 

\2am^nY(f 

4  m^K^p^cf 


42  ALGEBRA. 

23.  (4 a^hz''  X  10 a^hh)  ^  5 a%W 

24.  (21a:y2«  --  3a;y'z)(-  2a;y2). 

25.  104 a5V  ^  (91  a^5V  ^  7 a*6^:r). 

26.  (24 a^^»^^  -f-  3 a^h-")  +  (35 a«^» V  ^  -  5 a^hx). 

27.  85a^"*+^^5a^'^-"\  28.    84^"-*-^  12a^ 

60.   To  divide  One  Polynomial  by  Another. 

If  the  divisor  (one  factor)  =  a-\-h-\-  c, 

and  the  quotient  (other  factor)      =  ■^+P  +  $', 

an  +  hn  +  en 
then  the  dividend  (product)  =  <  -\-a'p  -\-hp-^  cp 

^-{-aq-\-hq-\-cq. 

The  first  term  of  the  dividend  is  an ;  that  is,  the  product 
of  a,  the  first  term  of  the  divisor,  by  n,  the  first  term  of 
the  quotient.  The  first  term  n  of  the  quotient  is  therefore 
found  by  dividing  an,  the  first  term  of  the  dividend,  by  a, 
the  first  term  of  the  divisor. 

If  the  partial  product  formed  by  multiplying  the  entire 
divisor  by  n  is  subtracted  from  the  dividend,  the  first  term 
of  the  remainder  ap  is  the  product  of  a,  the  first  term  of 
the  divisor,  by  p,  the  second  term  of  the  quotient ;  that  is,, 
the  second  term  of  the  quotient  is  obtained  by  dividing  the 
first  term  of  the  remainder  by  the  first  term  of  the  divisor. 
In  like  manner,  the  third  term  of  the  quotient  is  obtained 
by  dividing  the  first  term  of  the  new  remainder  by  the 
first  term  of  the  divisor ;  and  so  on.  Hence  we  have  the 
following  rule : 

Arrange  both  the  dividend  and  divisor  in  ascending  or 
descending  powers  of  some  common  letter. 

Divide  the  first  term  of  the  dividend  hy  the  first  term  of 
the  divisor. 


DIVISION.  43 

Write  the  result  as  the  first  term  of  the  quotient. 

Multiply  all  the  terms  of  the  divisor  hy  the  first  term  of 
the  quotient. 

Subtract  the  product  from  the  dividend. 

If  there  he  a  remainder,  consider  it  as  a  new  dividend 
and  proceed  as  before. 

61.  It  is  of  fundamental  importance  to  arrange  the  divi- 
dend and  divisor  in  the  same  order  with  respect  to  a  com- 
mon letter,  and  to  keep  this  order  throughout  the  operation. 

The  beginner  should  study  carefully  the  processes  in  the 
following  examples : 

(1)   Divide  a;'  -f  18a;  +  77  by  x  -f  7. 


:r^  +  18:r-f  77 
x''-\'    Ix 


x+    7 


a; +11 


lla;+77 
lla;+77 


Note.  The  student  will  notice  that  by  this  process  we  have  in 
effect  separated  the  dividend  into  two  parts,  x^  +  7x  and  11a;  +  77, 
and  divided  each  part  by  x  +  1,  and  that  the  complete  quotient  is 
the  sum  of  the  partial  quotients  x  and  11.     Thus, 

x2  +  18a;  +  77  =  a;2  +  7a;  +  lla  +  77  =  (a;^  +  7a;)  +  (llx  +  77); 

.    a;2  +  18a;  +  77      a;'+7a;  ,  11  a; +  77      ^^-,-, 
•  •  t; == ^  + f7~  =  a;  +  11. 

x  +  7  a;  +  7  a;  +  7 

(2)  Divide  4  aV  —  4  aV  -}-x^  —  a^  by  x^  —  al 
Arrange  according  to  descending  powers  of  x. 


rr«-4aV  +  4aV-a« 

x'~    a' 

x'-    aV 

a;*-3aV  +  a* 

-3aV  +  4aV-a*' 
-3aV  +  3«V 

a  V  -  a' 

44 


ALGEBRA. 


(3)  Divide  22a'Z>'  +  155*  +  3  a*  -  10  a'b  --  22  ab' 

hj  a''  +  Sb'-2ab. 
Arrange  according  to  descending  powers  of  a. 

3a*  -  lOa'b  +  22a'b'  -  22ab'  +  15^>*|    a'-2ab-^Sb' 


3a*-    6a'b+    9a'b'                             3a'- 

-4tab  +  6b^ 

-  ^a'b  +  lSa'b'-22ab' 

-  4a'b-\-    8a'b'~  12  ab' 

bd'b'-10ab'-\-15b' 
^n:^b'-^10a¥+15b* 

(4)  Divide  bx'-x-i-l-  3x'  by  1  +  3^^  -  2x. 
Arrange  according  to  ascending  powers  of  x. 


1—    x-{-5x^ 
l-2.r  +  3:i^2 


3^'^ 


1-2^:4-3:1;'' 


1+     X 
-Sx* 


x-Sx'  +  5x' 
x-2x''  +  Sx' 

—  x'  +  2x'  —  Sx' 

-  x'-]-2x'-Sx' 


(5)  Divide  x^  +  y^  -^  z^  —  Zxyz  by  a;  +  y  +  2. 
Arrange  according  to  descending  powers  of  x. 

x^  —  ?>yzx  +  ^Sd-  z^\x-\-y  +  z 

x'  +  i/x"  +  z: 


—  yx  —  zx 


yx-zx-\-  y^ 

+  y'  +  ^ 


yz  +  z" 


—  zoi^  +    y^x  —  2yzx  +  3/^  +  2^ 

—  zx"^  —    yza;  —  z^x 


y'^x  —    y2;.'r  +  ^^^  +  y*  +  2' 

?/a: +  y^  +  y'g 

—    yz^r  +  z'^o;  —  y'^z  +  2;' 


yza; 


y'z  -  yz' 


z^x  -f  yz^  +  2^ 
2^^  +  yz'  +  2^ 


DIVISION.  45 

Exercise  17. 
Divide 

1.  a^~7x+l2hy  x-B. 

2.  x''  +  x-12hy  x  +  9. 

3.  2a;3-:c2+3a;-9by  2a:-3. 

4.  6a^-{-Ux'-4:X  +  24:hy  2x  +  6. 

5.  Sx'-i-xi-9x'-lhjSx-l. 

6.  7^^^  +  58a;-24a,-2-21by  7a;-3. 

7.  a;^  — 1  by  a;  — 1. 

8.  a^-2ab'-}-b'hj  a-b. 

9.  a;* -Sly*  by  a;- 3y. 

10.  x^  —  y^  hy  X  —  y. 

11.  a^  +  325^by  a  +  2^>. 

12.  2a*  +  27a63-815*by  a  +  36. 

13.  :^*  +  lla;^-12a;-5a;'  +  6by  34-a;'-3a;. 

14.  x'-9x^-i-x^-16x-4thy  x''-{-4:i-4:X. 

15.  36  +  a;*-13a;Uy  6  +  a;'  +  5a:. 

16.  a;*  +  64bya;\+4a:4-8. 

17.  a;*  +  a;'  +  57-35a;-24a;'^bya;^-3  +  2a;. 

18.  l-x  —  Bx''-x^hyl-{-2x  +  x\ 

19.  :c«-2a;'+l  by  a;'  — 2a;+l. 

20.  a*  +  2a^6"^  +  9Z>*by  a''-2a6  4-36V 

21.  4a;^— a;3  +  4a;by  2  +  2a;'  +  3a;. 

22.  a^- 243  by  a -3. 

23.  18a;*  +  82a;^  +  40-67a;-45a;^by  3a;'  +  5-4a7. 

24.  x'-Qxy-dx'-y'hy  x'  +  i/-{-Sx. 


46  ALGEBRA. 

25.  x'^^xhf  -^xhj-^y'hy  x^-2,xy-\-2y\ 

26.  x^-\-x'^y'^-\-y^hY  x^  —  xy -\-y^. 

27.  x'-]-x^-\-x*y  +  f  —  ^^xy"^  —  x^y""  hj  s? -\- x  —  y . 

28.  2x'  —  Sy^  -[-xy  —  xz  —  ^yz  —  z'  by  2^;  +  3?/  +  z. 

29.  12+82a;^  +  106a;*-70a;^-112a;^-38a; 

by  2>-bx-\-lx\ 

30.  a;^  +  y^  by  a;*  —  x^y  +  a;Y^  —  xy^  +  y*. 

31.  2a;*  +  2a;y-2a;3/3-7a;V-y*by  2a;'4-/  — ^2/. 

32.  l^x'  +  4a;y  +  3/*  by  43;^^-  2a;y  +  y\ 

33.  32a^^»  +  8a'Z)'-a6^-4a^i*-56a*6^ 

by  ^'-4a'6  +  6a6l 

34.  l-\-b3^-^x^hyl-x-{-^x\ 

35.  1  -  52a*5*  -  51  a'5^  by  4 a'6^  +  3  a6  -  1. 

36.  a;''?/  —  xy''  by  a;^y+  2x^  —  2a;y  —  y*. 

37.  a;«  +  IbxY  +  15:ry  +  y'  -  ^^V  -  ^^V'  -  ^^^Y 

hj  x^-Sx''yi-Sxy^-y\ 

38.  a'  +  2a'b*-2a*b'~2a'b-6a'b'-Sab^ 

hy  a'- 2a'b-ab\ 

39.  Slx'y  +  ISa^y  -  54a;y  -  ISx'y* -  18xy'  -  9?/' 

by  3a;*  +  :ry  +  y*. 

40.  a'  +  2a'b  +  8a'b'  +  Sab'-i-16b'hja'  +  U\ 

41.  8/-a;«  +  21a;y-24:r2/^by  3a;?/  — a;'-y'. 

42.  16a*  +  9i*  +  8a'6^by  4a'  +  36'-4a5. 

43.  a^H-6'  +  c'  — 3a5cby  a  +  5  +  c. 

44.  a^  +  8b^-{-c^-Qabchy  a'-{-4:b'  +  c''-ac-2ab  —  2bc. 

45.  a-'  +  6'4-c'  +  3a'6  +  3a6'by  a  +  6  +  c. 


DIVISION.  47 

62.   The  operation  of  division  may  be  shortened  in  some 
cases  by  the  use  of  parentheses.     Thus  : 

x^-\-(a  -\- b -\- c) x^ -\-  {ah  -{-ac-\-  he)  x-\-abc\x  -f-  h 


^•'  +  (    +^       )^'  x'  +  {a-\-c)x-\-ac 

{a        -\-  c)  x'^  -f  {ah  -j-  ac  +  he)  x 
(a        -{-c)ar^  +  {ah         +  be)  x 


aex  +  o,bc 

aex  +  ahe 


Exercise  18. 
Divide 

1.  a\h  +  c)  +  ^\a  -  c)  +  e\a  -b)^  ahe\^y  a^h -\- c. 

2.  y?  —  {a-\-h  -\-  c)x^  ■\-  {ah  -\- ae -[■  he) x  —  ahe 

by  x^  —  {a -{- h)  X -\-  ah. 

3.  x^  -  2ax^  -I-  (a'  -^  ah  -  b'')x  -  a^h  +  ah""  hy  x-a-\-h. 

4.  X*'  —  (a^  —  5  —  c)  x^  —  {h  —  e)  ax  -f-  5^  by  x^  —  ax  -J-  c. 

5 .  y^  —  (wi  +  w +/>)  y^  +  ( wn  +  mp  -f  wp)  y  —  mnp  by  y  —  jo. 

6.  x'  -^  {b  -\-  a)a?  -  {^  -  ba-^h)x''  -  {^a-^bh)x  +  U 

hj  x"^  -\-bx  —  ^. 

7.  .^*-(a+5  +  c  +  c?)a;'  +  (a5  +  ac  +  acZ+^^  +  ^c?+cc^)a:2 

—  {ahe  -\-ahd  +  a<?c?  +  ^cd)  ^  +  a^<?c? 
by  rc^  —  (a  +  c)  a:  +  ae. 

8.  o;^  —  (m  —  <?)a;*  +  (w  —  cm  +  cf )a:'  +  (^  +  ^^  —  c???i) a;' 

+  {er  +  c?n)  a:  +  c?r  by  ar*  —  ma?  -{-nx-\-  r. 

9.  rr^  —  mx*  +  ^^^  —  '^^^^  +  ^^  —  1  by  a;  —  1. 

10.   {x  +  yy  +  S{x-{-y)'z  +  S{x  +  i/)z'  +  z' 
hj{x-]-y)'  +  2{x-\-7/)z  +  z\ 


CHAPTER  V. 

SIMPLE   EQUATIONS. 

63.  Equations.  An  equation  is  a  statement  in  symbols 
that  two  expressions  stand  for  the  same  number.  Thus, 
the  equation  3:c  -j-  2  =  8  states  that  Sx-{-2  and  8  stand  for 
the  same  number. 

64.  That  part  of  the  equation  whicli  precedes  the  sign 
of  equality  is  called  the  first  member,  or  left  side,  and  that 
which  follows  the  sign  of  equality  is  called  the  second  mem- 
ber, or  right  side. 

65.  The  statement  of  equality  between  two  algebraic 
expressions,  if  true  for  all  values  of  the  letters  involved, 
is  called  an  identical  equation;  but  if  true  only  for  certain 
particular  values  of  the  letters  involved,  it  is  called  an 
equation  of  condition.  Thus,  (a  -{-bf  =  a^  -{~2ab-\-  b^,  which 
is  true  for  all  values  of  a  and  b,  is  an  identical  equation  ; 
and  Sx-\-2  =  8,  which  is  true  only  when  x  stands  for  2,  is 
an  equation  of  condition. 

For  brevity,  an  identical  equation  is  called  an  identity, 
and  an  equation  of  condition  is  called  simply  an  equation. 

66.  We  often  employ  an  equation  to  discover  an  unknown 
number  from  its  relation  to  known  numbers.  We  usually 
represent  the  unknown  number  by  one  of  the  last  letters  of 
the  alphabet,  as  x,  y,  z;  and,  by  way  of  distinction,  we  use 
the  first  letters,  a,  5,  <?,  etc.,  to  represent  numbers  that  are 
supposed  to  be  known,  though  not  expressed  in  the  number- 


SIMPLE   EQUATIONS.  49 

symbols  of  Arithmetic.  Thus,  in  the  equation  ax-\-b  =  c, 
X  is  supposed  to  represent  an  unknown  number,  and  a,  b, 
and  c  are  supposed  to  represent  known  numbers. 

67.  Simple  Equations.  An  integral  equation  which  con- 
tains the  first  power  of  the  symbol  for  the  unknown  number, 
X,  and  no  higher  power,  is  called  a  simple  equation,  or  an 
equation  of  the  first  degree.  Thus,  ax-\-h  =  c  is  a  simple 
equation,  or  an  equation  of  the  first  degree  in.  x. 

68.  Solution  of  an  Equation.  To  solve  an  equation  is  to 
find  the  unknown  numbei*;  that  is,  the  number  which,  when 
substituted  for  its  symbol  in  the  given  equation,  renders 
the  equation  an  identity.  This  number  is  said  to  satisfy 
the  equation,  and  is  called  the  root  of  the  equation. 

69.  Axioms.  In  solving  an  equation,  we  make  use  of 
the  following  axioms : 

Ax.  1.  If  equal  numbers  be  added  to  equal  numbers, 
the  sums  will  be  equal. 

Ax.  2.  If  equal  numbers  be  subtracted  from  equal  num- 
bers, the  remainders  will  be  equal. 

Ax.  3.  If  equal  numbers  be  multiplied  by  equal  numbers, 
the  products  will  be  equal. 

Ax.  4.  If  equal  numbers  be  divided  by  equal  numbers, 
the  quotients  will  be  equal. 

If,  therefore,  the  two  sides  of  an  equation  he  increased  by, 
diminished  by,  multiplied  by,  or  divided  by  equal  numbers, 
the  results  will  be  equal. 

Thus,  if  8a;  =  24,  then  8a;  +  4  =  24  +  4,  8a:  -  4  - 24  -  4,, 
4  X  8a;  =  4  X  24,  and  8a;  ^  4  =  24  ^  4. 

70.  Transposition  of  Terms.  It  becomes  necessary  in  solv- 
ing an  equation  to  bring  all  the  terms  that  contain  the 


50  ALGEBRA. 

symbol  for  the  unknown  number  to  one  side  of  the  equation, 
and  all  the  other  terms  to  the  other  side.  This  is  called 
transposing  the  terms.     We  will  illustrate  by  examples : 

(1)  Find  the  number  for  which  x  stands  when 

16a;-ll-7a:>70. 
First  subtract  1  x  from  both  sides  (Ax.  2),  which  gives 

9:r-ll=70. 
Then  add  11  to  these  equals  (Ax.  1),  which  gives 

9:r  =  81. 
Divide  both  sides  by  9  (Ax.  4), 
x  =  9. 

(2)  Find  the  number  for  which  x  stands  when  x-\-h  =  a. 
The  equation  is  x-]-h—-o.. 

Subtract  h  from  each  side,  x-\-h  —  h  =  a  —  h.       (Ax.  2) 
Since  +  h  and  —bm  the  left  side  cancel  each  other 
(§  14),  we  have  x^  a  —  h. 

(3)  Find  the  number  for  which  x  stands  when  x  —  h  =  a. 
The  equation  is  x  —  h^a. 

Add  +  5  to  each  side,         x  —  b  -\-h  =  a-\-h.       (Ax.  1) 
Since  —b  and  -\-b  in  the  left  side  cancel  each  other 
(§  14),  we  have  x  =  a-\-b. 

71.  The  effect  of  the  operation  in  the  preceding  equa- 
tions, when  Axioms  (1)  and  (2)  are  used,  is  to  take  a  term 
from  one  side  and  to  put  it  on  the  other  side  with  its  sign 
changed.  We  can  proceed  in  a  like  manner  in  any  other 
case.     Hence  the  general  rule : 

72.  Any  term  may  be  transposed  from  one  side  of  an 
equation  to  the  other  provided  its  sign  is  changed. 


SIMPLE   EQUATIONS.  ^       51 

73.  Any  term,  therefore,  which  occurs  on  both  sides  with 
the  same  sign  may  be  removed  from  both  without  affecting 
the  equality. 

74.  The  sign  of  every  term  of  an  equation  may  be 
changed,  for  this  is  effected  by  multiplying  by  —  1,  which 
by  Ax.  3  does  not  destroy  the  equality. 

75.  Verification.  When  the  root  is  substituted  for  its 
symbol  in  the  given  equation,  and  the  equation  reduces  to 
an  identity,  the  root  is  said  to  be  verified. 

(1)  What  number  added  to  twice  itself  gives  24  ? 

Let  X  stand  for  the  number ; 
then  2x  will  stand  for  twice  the  number, 
and  the  number  added  to  twice  itself  will  hQ  x  -\-2x. 

But  the  number  added  to  twice  itself  is  24 ; 

.•.a;  +  2a:  =  24. 

Combining  x  and  2x,  3a;  —  24. 

Divide  by  3,  the  coefiicient  of  rr,      x  =  ^.  (Ax.  4) 

Verification.  :r  +  2rc  =  24, 

8  +  2  X  8  =  24, 

8  +  16  =  24, 

24  =  24. 

(2)  If  4  a;  —  5  =  19,  for  what  number  does  x  stand  ? 

We  have  the  equation  4  a;  —  5  =  19. 

Transpose  —  5  to  the  right  side,    4a;  =  19  -f  5. 

Combine,  4  a;  =  24. 

Divide  by  4,  x  =  6.  (Ax.  4) 

Verification.  4  a;  —  5  =  19, 

4x6-5  =  19, 

24  -  5  =  19, 

19  =  19. 


52      •  ALGEBRA. 

(3)  If  3a;— 7  stands  for  the  same  number  as  14  — 4a;, 
what  number  does  x  stand  for  ? 

We  have  the  equation 

3a;- 7  =  14 -4a;. 
Transpose  4  a;  to  the  left  side,  and  7  to  the  right  side, 

3a;  +  4a;  =  14 +  7. 
Combine,  7  a;  =  21. 

Divide  by  7,  a;  =  3. 

Verification.      3a;— 7  =  14  — 4a;, 

3x3-7  =  14-4x3, 

2  =  2. 

(4)  Solve  the  equation  (a;  -  3)  (a;  —  4)  =  a;  (a;  —  1)  —  30. 
Wehave    (a;  -  3)  (a;  -  4)  =  rir  (a;  -  1)  -  30. 

Remove  the  parentheses, 

a;'-7a;+12  =  a;'-a;-30. 

Since  x^  on  the  left  and  a;^  on  the  right  are  precisely  the 
same,  including  the  sign,  they  may  be  cancelled. 

Then  -  7a;+ 12  =  -  a;  -  30. 

Transpose  —  a;  to  the  left  side,  and  +  12  to  the  right  side, 

-7a;  +  ^  =  -30-12. 
Combine,  —  6  a;  =  —  42. 

Divide  by  —  G,  a;  =  7. 

Verification. 

(7 -3)  (7 -4)  =  7  (7-1) -30, 
4x3  =  7x6- 30, 
12  =  42  -30, 
12  =  12. 


SIMPLE   EQUATIONS.  63 

76.  Hence,  to  solve  an  equation  with  one  unknown 
number, 

Transpose  all  the  terms  involving  the  unknown  number 
to  the  left  side,  and  all  the  other  terms  to  the  right  side; 
combine  the  like  terms,  and  divide  both  sides  by  the  coefficient 
of  the  unknown  number. 

Exercise  19. 
Find  the  value  of  x  in 

1.  5:r-l  =  19.  8.  lGa;-ll  =  7rr-f  70. 

2.  3x+6  =  12.  9.  24a: -49  =  19^7 -14. 

3.  24:r=7:r  +  34.  10.  3a:  +  23  =  78 -2a:. 

4.  8a: -29  =  26 -3a:.  11.  26  -  8a:  =  80- 14a:. 
6.    12-5a:  =  19  — 12a:.  12.  13  -  3a:  =  5a7- 3. 

6.  3a:  +  6-2a:=7a:.  13.    3a:- 22  =  7a:  +  6. 

7.  5a:  +  50  =  4a:  +  56.         14.    8  +  4a:  =  12a:  -  16. 

15.  5a: -(3a: -7) -4a: -(6a: -35). 

16.  6 a:- 2(9 -4a:) +3(5a:- 7) -10a:- (4  + 16a: 4- 35). 

17.  9a:-3(5a:-6)  +  30  =  0. 

18.  a:  -  7  (4a:  -  11)  =-  14  (a:  -  5)  -  19  (8  -  a:)  -  61. 

19.  (a:  +  7)(a:-3)  =  (a:-5)(ar-15). 

20.  (a:-8)(a:+12)-=(:r+l)(a:-6). 

21 .  (a:  -  2)(7  -  a:)  +  (a:  -  5)(a:  +  3)  -  2  (a:  -  1)  + 12  =  0. 

22.  (2a:-7)(a:+5)  =  (9-2a:)(4-a:)  +  229. 

23.  14-a:-5(a:  +  3)(a:-h2)  +  (5-a:)(4-5a:)  =  45ar-76. 

24.  (a:  +  5)'^-(4-a:)^-21a:. 

25.  5(a:  -  2)^  +  7 (a:  -  3)^  =  (3a:  -  7)(4a:  -  19)  +  42. 


54  ALGEBRA. 

77.  Statement  and  Solution  of  Problems.  The  difficulties 
which  the  beginner  usually  meets  in  stating  problems  will 
be  quickly  overcome  if  he  will  observe  the  following  direc- 
tions: 

Study  the  problem  until  you  clearly  understand  its  mean- 
ing and  just  what  is  required  to  be  found. 

Remember  that  x  must  not  be  put  for  money,  length, 
time,  weight,  etc.,  but  for  the  required  number  of  specified 
units  of  money,  length,  time,  weight,  etc. 

Express  each  statement  carefully  in  algebraic  language, 
and  write  out  in  words  what  each  expression  stands  for. 

Do  not  attempt  to  form  the  equation  until  all  the  state- 
ments are  made  in  symbols. 

We  will  illustrate  by  examples  : 

(1)  John  has  three  times  as  many  oranges  as  James,  and 
they  together  have  32.     How  many  has  each  ? 

Let        X  be  the  number  of  oranges  James  has  ; 
then         3x  is  the  number  of  oranges  John  has  ; 
and   .T  +  3  a;  is  the  number  of  oranges  they  together  have. 

But  32  is  the  number  of  oranges  they  together  have  ; 
.-.  a;  +  3x  =  32; 
or,  4  a;  =  32, 

and  x  =  S. 

Since  a;  =  8,  3a;  =  24. 

Therefore  James  has  8  oranges,  and  John  has  24  oranges. 

Note.  Beginners  in  stating  the  preceding  problem  generally  write : 

Let  X  =  what  James  had. 

Now,  we  know  what  James  had.  He  had  oranges,  and  we  are  to 
discover  simply  the  number  of  oranges  he  had. 

(2)  James  and  John  together  have  $24,  and  James  has 
$8  more  than  John.     How  many  dollars  has  each  ? 


SIMPLE    EQUATIONS.  55 

Let  X  be  the  number  of  dollars  John  has ; 

then  X  +  8  is  the  number  of  dollars  James  has  ; 

and  X  +  (x  +  8)  is  the  number  of  dollars  they  both  have. 

But  24  is  the  number  of  dollars  they  both  have ; 
.-.  X  +  (x  +  8)  =  24  ; 
or,  X  +  X  +  8  =  24. 

Transpose  and  combine,      2x  =  16. 
Divide  by  2,  x  =  8. 

Since  x  =  8,  x  +  8  =  16. 

Therefore  John  has  $8,  and  James  has  $16. 

Note.   The  beginner  must  avoid  tlfb  mistake  of  writing 

Let  X  =  John's  money. 

We  are  required  to  find  the  number  of  dollars  John  has,  and  there- 
fore X  must  represent  this  required  number. 

(3)  A  and  B  had  equal  sums  of  money;  B  gave  A  $5, 
and  then  3  times  A's  money  was  equal  to  11  times  B's 
money.     What  had  each  at  first  ? 

Let  X  =  number  of  dollars  each  had  ; 

then  X  +  5  =  number  of  dollars  A  had  after  receiving  f  5 ; 

and  X  —  5  =  number  of  dollars  B  had  after  giving  A  $5. 

Since  3  times  A's  money  is  now  equal  to  11  times  B's,  we  have 
therefore  the  equation  : 

3  (x  +  5)  =  11  (x  -  5). 

Removing  parentheses,    3x  +  15  =  11  x  —  55. 

Transposing,  3  x  —  11  x  =  —  55  —  15. 

Collecting,  -8x  =  -70. 

Dividing  by  —  8,  x  =  8|. 

Therefore,  each  had  $8.75. 


56  ALGEBKA. 

(4)  Find  a  number  whose  treble  exceeds  50  by  as  much 
as  its  double  falls  short  of  40. 

Let  X  =  the  number ; 

then  3  X  =  its  treble ; 

and  3  x*  —  50  =  the  excess  of  its  treble  over  50  ; 

also,  40  —  2  a;  =  the  number  its  double  lacks  of  40, 

Since  the  excess  of  3  a;  over  50  equals  the  number  2x  lacks  of  40, 
we  have 

3  a; -50  =  40 -2a;; 

3a;  +  2a;  =  40 +  50; 

5a;  =  90; 

a;  =18. 

Therefore  the  number  required  is  18. 

(5)  Find  a  number  that  exceeds  50  by  10  more  than  it 


falls  short  of  80. 

Let 

x  = 

=  the 

required  number ; 

then 

X  - 

-50  = 

=  its  ( 

sxcess  over  50 ; 

and 

80 

—  x  = 

-the 

number  it  lacks  of  80. 

Hence, 

a;-50- 

-(80- 

-a;)  = 

=  the  excess. 

But 

10  = 

=  the 

excess. 

•.  a;-50- 

-(80- 

-a;)  = 

=  10, 

or 

a;-50 

-80 

+  x  = 
2a;  = 

=  10. 
=  140 

and 

X  = 

=  70. 

Therefore  the  number  required  is  70. 

Exercise  20. 

1 .  To  the  double  of  a  certain  number  I  add  14,  and  obtain 

as  a  result  154.     What  is  the  number? 

2.  To  four  times  a  certain  number  I  add  16,  and  obtain 

as  a  result  188.     What  is  the  number  ? 

3.  By  adding  46  to  a  certain  number,  I  obtain  as  a  result 

a  number  three  times  as  large  as  the  original  num- 
ber.    Find  the  original  number. 


SIMPLE   EQUATIONS.  57 

4.  One  number  is  three  times  as  large  as  another.     If  I 

take  the  smaller  from  16  and  the  greater  from  30, 
the  remainders  are  equal.     What  are  the  numbers? 

5.  Divide  the  number  92  into  four  parts,  such  that  the 

first  exceeds  the  second  by  10,  the  third  by  18,  and 
the  fourth  by  24. 

6.  The  sum  of  two  numbers  is  20 ;  and  if  three  times  the 

smaller  number  is  added  to  five  times  the  greater, 
the  sum  is  84.     What  are  the  numbers  ? 

7.  The  joint  ages  of  a  father  and  son  are  80  years.    If  the 

age  of  the  son  were  doubled,  he  would  be  10  years 
older  than  his  father.     What  is  the  age  of  each  ? 

8.  A  man  has  6  sons,  each  4  years  older  than  the  next 

younger.  The  eldest  is  three  times  as  old  as  the 
youngest.     What  is  the  age  of  each  ? 

9.  Add  $24  to  a  certain  sum,  and  the  amount  will  be  as 

much  above  |80  as  the  sum  is  below  $80.     What 
•   is  the  sum  ? 

10.  Thirty  yards  of  cloth  and  40  yards  of  silk  together  cost 

$  330 ;  and  the  silk  costs  twice  as  much  a  yard  as  the 
cloth.     How  much  does  each  cost  a  yard  ? 

11.  Find  the  number  whose  double  increased  by  24  exceeds 

80  by  as  much  as  the  number  itself  is  less  than  100. 

12.  The  sum  of  $600  is  divided  among  A,  B,  0,  and  D. 

A  and  B  have  together  $280,  A  and  0  $260,  and 
A  and  D  $  220.     How  much  does  each  receive  ? 

13.  In  a  company  of  266  persons  composed  of  men,  women, 

and  children,  there  are  twice  as  many  men  as  women, 
and  twice  as  many  women  as  children.  How  many 
are  there  of  each  ? 


58  ALGEBRA. 

14.  Find  two  numbers  differing  by  8,  such  that  four  times 

the  less  may  exceed  twice  the  greater  by  10. 

15.  A  is  68  years  older  than  B,  and  A's  age  is  as  much 

above  60  as  B's  age  is  below  50.  Find  the  age  of 
each. 

16.  A  man  leaves  his  property,  amounting  to  $7500,  to  be 

divided  among  his  wife,  his  two  sons,  and  three 
daughters,  as  follows:  a  son  is  to  have  twice  as 
much  as  a  daughter,  and  the  wife  $  500  more  than 
all  the  children  together.  How  much  will  be  the 
share  of  each  ? 

17.  A  vessel  containing  some  water  was  filled  by  pouring 

in  42  gallons,  and  there  was  then  in  the  vessel  seven 
times  as  much  as  at  first.  How  much  did  the  vessel 
hold? 

18.  A  has  1 72  and  B  has  $  52.     B  gives  A  a  certain  sum ; 

then  A  has  three  times  as  much  as  B.  How  much 
did  A  receive  from  B  ? 

19.  Divide  90  into  two  such  parts  that  four  times  one 

part  may  be  equal  to  five  times  the  other. 

20.  Divide  60  into  two  such  parts  that  one  part  exceeds 

the  other  by  24. 

21.  Divide  84  into  two  such  parts  that  one  part  may  be 

less  than  the  other  by  36. 

Note.  When  we  have  to  compare  the  ages  of  two  persons  at  a 
given  time,  and  also  a  number  of  years  after  or  before  the  given 
time,  we  must  remember  that  both  persons  will  be  so  many  years 
older  or  younger. 

Thus,  if  X  represent  A's  age,  and  2x  B's  age,  at  the  present  time, 
A's  age  five  years  ago  will  be  represented  by  a;  —  5 ;  and  B's  by 
2  a;  —  5.  A's  age  five  years  hence  will  be  represented  by  a;  +  5 ;  and 
B's  age  by  2x  +  5. 


SIMPLE   EQUATIONS.  '  59 

22.  A  is  twice  as  old  as  B,  and  22  years  ago  he  was  three 

times  as  old  as  B.     What  is  A's  age? 

23.  A  father  is  30  and  his  son  6  years  old.     In  how  many 

years  will  the  father  be  just  twice  as  old  as  the  son  ? 

24.  A  is  twice  as  old  as  B,  and  20  years  since  he  was  three 

times  as  old.     What  is  B's  age  ? 

25.  A  is  three  times  as  old  as  B,  and  19  years  hence  he 

will  be  only  twice  as  old  as  B.  What  is  the  age 
of  each  ? 

26.  A  man  has  three  nephews  ;  his  age  is  50,  and  the  joint 

ages  of  the  nephews  is  42.  How  long  will  it  be 
before  the  joint  ages  of  the  nephews  will  be  equal  to 
that  of  the  uncle  ? 

Note.  In  problems  involving  quantities  of  the  same  kind 
expressed  in  different  units,  we  must  be  careful  to  reduce  all  the 
quantities  to  the  same  unit. 

Thus,  if  X  denote  a  number  of  inches,  all  the  quantities  of  the  same 
kind  involved  in  the  problem  must  be  reduced  to  inches. 

27.  A  sum  of  money  consists  of  dollars  and  twenty-five-cent 

pieces,  and  amounts  to  $20.  The  number  of  coins 
is  50.     How  many  are  there  of  each  sort  ? 

28.  A  person  bought  30  pounds  of  sugar  of  two  different 

kinds,  and  paid  for  the  whole  $2.94.  The  better 
kind  cost  10  cents  a  pound  and  the  poorer  kind  7 
cents  a  pound.  How  many  pounds  were  there  of 
each  kind  ? 

29.  A  workman  was  hired  for  40  days,  at  $  1  for  every  day 

he  worked,  but  with  the  condition  that  for  every 
day  he  did  not  work  he  was  to  pay  45  cents  for  his 
board.  At  the  end  of  the  time  he  received  $22.60. 
How  many  days  did  he  work  ? 


60  ALGEBRA. 

30.  A  wine  merchant  has  two  kinds  of  wine  ;  one  worth  50 

cents  a  quart,  and  the  other  75  cents  a  quart.  From 
these  he  wishes  to  make  a  mixture  of  100  gallons, 
worth  $2.40  a  gallon.  How  many  gallons  must  he 
take  of  each  kind  ? 

31.  A  gentleman  gave  some  children  10  cents  each,  and 

had  a  dollar  left.  He  found  that  he  would  have 
required  one  dollar  more  to  enable  him  to  give  them 
15  cents  each.     How  many  children  were  there  ? 

32.  Two  casks  contain  equal  quantities  of  vinegar;  from 

the  first  cask  34  quarts  are  drawn,  from  the  second 
20  gallons ;  the  quantity  remaining  in  one  vessel  is 
now  twice  that  in  the  other.  How  much  did  each 
cask  contain  at  first  ? 

33.  A  gentleman  hired  a  man  for  12  months,  at  the  wages 

of  $90  and  a  suit  of  clothes.  At  the  end  of  7  months 
the  man  quits  his  service  and  receives  $33.75  and 
the  suit  of  clothes.  "What  was  the  price  of  the  suit 
of  clothes  ? 

34.  A  man  has  three  times  as   many   quarters   as   half- 

dollars,  four  times  as  many  dimes  as  quarters,  and 
twice  as  many  half-dimes  as  dimes.  The  whole  sum 
is  $7.30.     How  many  coins  has  he  in  all? 

35.  A  person  paid  a  bill  of  $15.25  with  quarters  and  half- 

dollars,  and  gave  51  pieces  of  money  in  all.  How 
many  of  each  kind  were  there  ? 

36.  A  bill  of  100  pounds  was  paid  with  guineas  (21  shil- 

lings) and  half-crowns  (2 J  shillings),  and  48  more 
half-crowns  than  guineas  were  used.  How  many  of 
each  were  paid? 


CHAPTER  VI. 
MULTIPLICATION   AND   DIVISION. 

Special  Rules. 

78.  Special  Kules  of  Multiplication.  Some  results  of  mul- 
tiplication are  of  so  great  utility  in  shortening  algebraic 
work  that  they  should  be  carefully  noticed  and  remem- 
bered.    The  following  are  important : 

79.  Square  of  the  Sum  of  Two  Numbers. 

(a4-67  =  (a  +  6)(a-f5) 

=  a(a-{-b)  +  b(a  +  b) 
=  a'  +  ab  +  ab  +  b' 
^a^-\-2ab^b\ 

Since  a  and  b  stand  for  any  two  numbers,  we  have 

Rule  1.  The  square  of  the  sum  of  two  numbers  is  the 
sum  of  their  squares  plus  twice  their  product. 

80.  Square  of  the  Difference  of  Two  Numbers. 

(a-by  =  (a-bXa-b) 

=  a(a  —  b)  —  b  (a  —  b) 

—  ^2  _  ^Z,  _  ^I      I      7,2 


=  a 


ab  —  ab-{-  b"^ 


=  a'~2ab  +  b\ 

Hence  we  have 

Rule  2.      The  square  of  the  difference  of  two  numbers  is 
the  sum  of  their  squares  yninus  twice  their  product. 


62  ALGEBRA. 

81.  Product  of  the  Sum  and  Difference  of  Two  Numbers. 

(a  +  b)(a  —  b)  =  a(a-b)  +  b(a-b) 
=  a^  —  ab  -]-  ab  —  b^ 
=  a'-b\ 
Hence  we  have 

Rule  3.  The  product  of  the  sum  and  difference  of  two 
)i  umbers  is  the  difference  of  their  squares. 

82.  Illustrations.    If  we  put  2  a;  for  a  and  3  for  ^,  we  have 
Rulel,  (2a; +  3)^  =  4x^+120; +  9. 

Rule  2,  (2x-Sy=4:x'-12x  +  9. 

Rule  3,  {2x  +  3)(2a;  -  3)  =  Ax'  -  9. 

If  we  are  required  to  multiply  a-\-b  -\-  c  by  a-i-b  —  c, 
we  may  abridge  the  ordinary  process  as  follows  : 

(a  +  5  +  c)(a  +  6-c)-[(a  +  5)  +  c][(a  +  5)-c] 
By  Rule  3,  =  («  +  hf  -  c' 

By  Rule  1,  =  a' -\-2ab  +  b' -  c\ 

If  we  are  required  to  multiply  a  +  J  —  cby  a  —  b-\-c,  we 
may  put  the  expressions  in  the  following  forms,  and  per- 
form the  operation  : 

(a  +  6  -  c)(a  -b  +  c)-=[a-\-{b~  c)][a  ~  {b  -  c)] 

By  Rule  3,  =  a'  -  (b  -  cf 

By  Rule  2,  =  a' -  (b'  -2bc  +  c') 

^a'-b'-\-2bc-c\ 


SPECIAL    RULES    OF   MULTIPLICATION.  63 

Exercise  21. 
Write  the  product  of 

1.  {x^y)\  9.  {ah^cdf. 

2.  {y  —  zf.  10.  (3mn  — 4)1 

3.  {2x-{-l)\  11.  (12  +  5x)l 

4.  (2a  +  55)^  12.  (4:ry'^-yz^)^ 
5.(l-a;7.  13.  (^abc-bcdy. 

6.  (3aa;-4rr7.  14.    (4a:^-a:2/7. 

7.  (l-7a)^  15.    (:t.  +  y)(:r-3/). 

8.  (5 ^2/ +  2)'.  16.    (2ai-b)(2a-b). 

17.  l  +  a  +  i  and  1  —  a  —  b. 

18.  1  — a-f  5  and  1 +  a  — 5. 

19.  a2  +  a5  +  5^  and  a2-a5  +  5^ 

20.  36X  +  2^>  — c  and  3a-2^>  +  c. 

83.   Square  of  any  Polynomial.     If  we  put  x  for  a,  and 

y  -f-  2;  for  5,  in  the  identity 

(a  +  by  =  a'  +  2ab-{-b\ 

we  shall  have 

[x  +  {y  +  z)J  =  x'  +  2x(y  +  z)-\-(i/  +  zy, 

or  (a:  +  2/  +  z)^  =  ^'^  +  2a:y  +  2  372  +  y'  -f-  2y2  +  z' 

=  ^'  +  y'  +  z'  +  2  ary  +  2  arz  4-  2 yz. 

It  will  be  seen  that  the  complete  product  consists  of  the 
sum  of  the  squares  of  the  terms  of  the  given  expression  and 
twice  the  products  of  each  term  into  all  the  terms  that  fol- 
low it. 


64  ALGEBRA. 

Again,  if  we  put  a  —  b  for  a,  and  c  —  d  for  5,  in  the  same 
identity, 

{a-\-hf  =  a^-\-'2.ab'\-h\ 

we  shall  have 

[{a-h)  +  {o-d)J 

=  {a-hy-^2{a-  h)  {c-d)-\-{c-  df 
=  (a'-2ab  +  b')  +  2a(c--d)-2b{c-d)-\-(c'-2cdi-d') 
=  a''-2ab  +  b'+2ac—2ad-2bc-{-2bd+c''-2cd-\-d'' 
=  a''  +  b^  +  c''-{-d'-2ab-\-2ac-2ad~2bc+2bd-2cd. 

Here  the  same  law  holds  as  before,  the  sign  of  each 
double  product  being  +  or  — ,  according  as  the  factors  com- 
posing it  have  like  or  unlike  signs.  The  same  is  true  for 
any  polynomial.     Hence  we  have 

EuLE  4.  The  square  of  a  polynomial  is  the  sum  of  the 
squares  of  the  several  terms  and  twice  the  products  obtained 
by  multiplying  each  term  into  all  the  terms  that  follow  it. 

Exercise  22. 

1.  {x-\-y^zf=  9.  {a'  +  b'  +  cy  = 

2.  {x--y-\-zy=  10.  {o(^-f-^y  = 

3.  {m  +  n~p-qY^  11.  {x -\-2y ~2>zf  = 

4.  (o;^  +  2rp  -  3)^  -  12.  {x'-27f-\-^zJ-= 

5.  (a:^-6.r+7)^=  13.  (x" -\- 2x -2)'-^ 

6.  {2x'-1x-\-^f^-  14.  {x^-bx  +  iy  = 

7.  (a;^  +  2/'-z7=  15.  {2x'' -^x  -  4:^  = 

8.  {x' -  ^a^y"" -\- yj -=r  16.  (x^2y  +  2>zy  = 

84,  Product  of  Two  Binomials.  The  product  of  two  bino- 
mials which  have  the  form  x-{-  a,x-{-b,  should  be  carefully 
noticed  and  remembered. 


(^-5)(T.- 

-'^)  =  x{x-^)-b{x-2>) 

=  x'-^x-bx  +  \b 

=  x'~Sx-{-lh. 

{x+b){x- 

-3)  =  :r(a;-3)  +  5(a:-3) 

=  ^'-3:r +  5:^-15 

=  x''  +  2x-~lb. 

SPECIAL    RULES    OF    MULTIPLICATION.  65 

(1)  (:r+  5)(a;  +  3)  =  x(:r  +  3)  +  5(:r+3) 

=  .r'  +  3a;  +  5a7+15 

(2) 


(3) 


(4)  {x  -  5)(:r  +  3)  =  x{x  +  3)  -  b{x  +  3) 

=  x'  -^?>x-bx-\b 
=  a;^-2.r-15. 

Each  of  these  results  has  three  terms. 

The  first  term  of  each  result  is  the  product  of  the  first 
terms  of  the  binomials. 

The  last  term  of  each  result  is  the  product  of  the  second 
terms  of  the  binomials. 

The  middle  term  of  each  result  has  for  a  coefficient  the 
algebraic  sum  of  the  second  terms  of  the  binomials. 

The  intermediate  step  given  above  may  be  omitted,  and 
the  products  written  at  once  by  inspection.     Thus, 

(1)  Multiply  a;  +  8  by  .T  +  7. 

8  +  7  =  15,    8x7  =  56. 
.-.  {x  +  K){x  +  7)  -  o:^  +  15^'  +  56. 

(2)  Multiply  X  -  8  by  a-  -  7. 

(-  8)  +  (-  7)  =  -  15,    (-  8)(-  7)  =  +  56. 
.-.  {x  -  8)(a;  -  7)  =  ^'  -  15ar  +  56. 


66 


ALGEBRA. 


(3)  Multiply  x—ly\)jx-{-^y. 

_  7  +  6  =  -  1,   (-  72/)  X  62/  =  -  423/2. 
.-.  {x  —  ly){x  -\-  Qy)  =  x^  —  xy  —  42y^ 

(4)  Multiply  x'  +  6  (a  +  5)  by  .t^  -  5  (a  +  5). 

+  6-5  =  1,   6(a  +  &)  X  -  5(a  +  6)  =  -  30(a  +  6)'^ 
.-.  [.^^+6  (a+h)][x'-5  (a+b)]  =  x'+{aib)  x^~^0 (ai b)\ 


Exercise  23. 


Write  the  product  of 

1.  (x  +  2Xx  +  3). 

2.  (x+lXx  +  b). 

3.  (x~~3)(x-6). 

4.  (.T  -  8X:r  -  1). 

5.  (:r  -  8)(:r  +  1). 

6.  (x-2)(x  +  5). 

7.  (.r-3X:r+7). 

8.  (x-2Xx-A). 

9.  (:r+ !)(:.  + 11). 
10.  (a:-2a)(a;~f-3a). 


11.  (x  —  cX^  —  d). 

12.  (a;-4y)(:r-f  y). 

13.  (a-2bXa-bb). 

14.  (x^  + 23/^X0;^ +  2/^). 

15.  (:r^-3a:yX^M-a;y). 

16.  (ax~9){ax  +  6). 

17.  (:r  +  a)(:u  -  Z)). 

18.  (:r- 11X^^4- 4). 

19.  (a: +12X^^-11). 

20.  (a: -10X^-5). 


85.    In  like  manner  the  product  of  any  two  binomials 
may  be  written.     Thus, 

(1)  {2a-bX^a-{-4:b)=.6a''-{-8ab-3ab~4b' 

=  Qa'  +  5ab-U\ 

(2)  (2x  +  3y)(3a:  -  2?/)  -  Qx'  -  4xy  +  9rry  -  63/' 


SPECIAL   RULES   OF  DIVISION.  67 

86.  Special  Eules  of  Division.  Some  results  in  division 
are  so  important  in  abridging  algebraic  work  that  they 
should  be  carefully  noticed  and  remembered. 

87.  Difference  of  Two  Squares. 
From  §  81,  (a  +  h){a  -b)  =  a'  -  b\ 

(^}  —  ip-  a^  —  h^ 

=  a  — 5,  and  —  =  a-f5.    Hence, 


Rule  1.     The  difference  of  the  squares  of  two  numbers  is 
divisible  by  the  sum,  and  by  the  difference,  of  the  numbers. 

Exercise  24. 

Write  the  quotient  of 

2.    1-=l4^.  10.    ^-^W. 

x^^y 


11. 


4.    — ^.  12. 


l-2a: 

\-^x' 

l  +  2a; 

9a^-b'' 

%a-\-b 

9a^-b^ 

Sa-b 

16a' -9b' 

4:a+Sb 

l-9z' 

1-30 

a'b'  -  c' 

ab  -\-  c 

4:x'  -  16y' 

l-(x-yy 
l-]-{x-y)' 

a'-jb  +  cf 
a  +  {b-\-c)' 

{x  +  yY-2b 
{x  +  y)  -  5 

l-{a-bbY 
\^{a-bb)' 

64-(2&4-3gV 
8 -(25  + 3c)  ■ 

8     ^ t:^  le     (a-hY-{c-d)' 

2x-{-^y  '    (a-b)^{c-d) 


13. 


14. 


15. 


68  ALGEBRA. 

88.  Sum  and  Difference  of  Two  Cubes.  By  performing  the 
division,  we  find  that 

9!-±±.=^a'-ab-{-b\  and ^^a'+ab  +  b'. 

a-\-  0  a  ~  b 

Hence, 

Rule  2.  The  swn  of  the  cubes  of  two  numbers  is  divisible 
by  the  sum  of  the  numbers,  and  the  quotient  is  the  sum  of 
the  sqiiares  of  the  numbers  minus  their  product. 

Rule  3.  The  difference  of  the  cubes  of  two  numbers  is 
divisible  by  the  difference  of  the  numbers,  and  the  quotient 
is  the  sum  of  the  squares  of  the  numbers  plus  their  product. 


Exercise  25. 
Write  the  quotient  of 

1.  {f-l)^{y-l).       „  8.  {^^a^b^-21x^)^{^ab-2>x). 

2.  {b'-\2b)^{b-b).  9.  (a^^f)-^{x-\-y). 

3.  (o,-^  -  216) -- (a  -  6).  10.  (l  +  8a')^(l  +  2a). 

4.  {x^-M^)^(x--l).  11.  (27a='  +  ^')^(3a+^). 

5.  (l-8.r^)-^(l-2.T).  12.  (8aV4-l)-(2aa;+l). 

6.  (8aV-l)^(2a.r-*l).  13.  {x'-\-21f)~{x-^?>y). 

7.  {l-21xY)~^{l~^xy).  14.  {b\2ci^f^z')-^{^xy-\-z). 

15.  (729a='  +  2165')-f-(9a  +  6Z'). 

16.  (64  a'' +  1000^') -^(4 a +  10 6). 

17.  (64a^5'  +  27a;')^(4a5  +  3a:). 

18.  (a;^  +  343)^(a:+7). 

19.  {21:(^y'^%z')^{?>xy-^2z). 


SPECIAL    RULES   OF   DIVISION.  69 

89.   Sum  and  Difference  of  any  Two  Like  Powers.     By  per- 
forming the  division,  we  find  that 


a  —  b 

a  +  b 
a'-b' 


a^  +  a'b-i-ab'  +  b'; 
a'  -  d'b  +  ab'  -  P  ; 


a  —  b 


+  b' 


=  a*  +  a'b-{-a'b'  +  ab'  +  b' 


a 


a'  -  a^b  +  a^b""  -  ab'  +  b\ 


a  +  b 
We  find  by  trial  that 

a"  +  b\  a*  4-  b\  a^  +  b\  and  so  on, 
are  not  divisible  by  a  +  5  or  by  a  —  b.     Hence, 
If  n  is  any  positive  integer, 

(1)  a"  +  5"  is  divisible  by  a -{-b  if  n  is  odd,  and  by  neither 
a-i-b  nor  a  —  b  if  n  is  even. 

(2)  a"  —  6"  is  divisible  by  a  —  b  if  n  is  odd,  and  by  both 
a  -f  ^  ctnd  a  —  b  if  n  is  even. 

Note.  It  is  important  to  notice  in  tlie  above  examples  that  the 
terms  of  the  quotient  are  all  positive  when  the  divisor  is  a  —  b,  and 
alternately  positive  and  negative  when  the  divisor  is  a  +  b;  also,  that 
the  quotient  is  homogeneous,  the  exponent  of  a  decreasing  and  of  b 
increasing  by  1  for  each  successive  terra. 

Exercise  26. 

Find  the  quotient  of 

1.  (x'-y')-^(x-y).  7.  (a' +  S2b') -^(a  +  2b). 

2.  {x'-y')-^(x  +  y).  8.  (a' -32b') -^  (a -2b). 

3.  (a'-x')^{a-x).  9.  (1  -  243a^)  ^  (1  -  3a). 

4.  {a^-x')-^(a-\-x).  10.  (243a^  + 1)  ^  (3a  + 1). 

5.  (81aV-lM3aa:+l).  11.  (x' -y')^  (x -y). 

6.  {Maf'-b')^{2a-b).  12.  (a^" -  1024) -  (a  +  2). 


CHAPTER  VII. 
FACTORS. 

90.  Eational  Expressions.  An  expression  is  o^ational  when 
none  of  its  terms  contain  square  or  other  roots. 

91.  Pactors  of  Eational  and  Integral  Expressions.  By  fac- 
tors of  a  given  integral  number  in  Arithmetic  we  mean 
integral  numbers  that  will  divide  the  given  number  with- 
out remainder.  Likewise  by  factors  of  a  rational  and  inte- 
gral expression  in  Algebra  we  mean  rational  and  integral 
expressions  that  will  divide  the  given  expression  without 
remainder. 

92.  Pactors  of  Monomials.  The  factors  of  a  monomial 
may  be  found  by  inspection.  Thus,  the  factors  of  l^a^h 
are  7,  2,  a,  a,  and  h. 

93.  Pactors  of  Polynomials.  The  form  of  a  polynomial 
that  can  be  resolved  into  factors  often  suggests  the  process 
of  finding  the  factors. 

Case  I, 

94.  When  all  the  Terms  have  a  Common  Pactor. 

(1)   Resolve  into  factors  2x'^  -{-  ^xy. 
Since  2:«;  is  a  factor  of  each  term,  we  have 

X ^^        j^^  =  x  +  ?>y. 

2x  2x       2x 

:.  2x'' -\-  ^xy  =  2x  {x  -]-?>y). 

Hence,  the  required  factors  are  2x  and  rr-f-  83/. 


FACTORS.  71 

« 

(2)   Resolve  into  factors  16  a^  +  4  a'  —  8a. 
Since  4a  is  a  factor  of  each  term,  we  have 

16a^  +  4a'-8a_16a^      4a'^      8a 
4a  4a        4a      4a 

=  4a'^  +  «-2. 
.-.  16a^  +  4a'-8a  =  4a(4a'  +  a-2). 

Hence  the  required  factors  are  4a  and  4:0^  -\-  a  — 2. 

Exercise  27. 

Resolve  into  factors : 

1.  5a' -15a.  4.   4:ar'7/ -12xy +  8xf. 

2.  6a' +  18a' -12a.  5.    y"  -  ay"  ^  by" -{- cy . 

3.  49:^2  _2U^_^  14.  e.    6a^Z>' -  21a*5' +  27a'6*. 

7.  54:r'y  +  108^y  -  243:iy- 

8.  45 xhf  -  90 a;y  -  360a;y . 

9.  70ay~140ay  +  210a?/^ 
10.    Z2a?}f'-\-ma^b^-V2&a^h\ 

Case  II. 

95.   When  the  Terms  can  he  grouped  so  as  to  show  a  Common 
Factor. 

(1)  Resolve  into  factors  ac-\-ad-{-hc-\-  bd. 

ac -{- ad -\- be -\- bd  =  {ac -\- ad) -\- {be -{- bd)  (1) 

=  a{c-^d)-\-b{c^d)  (2) 

=  {a-\-b){c-\-d).  (3) 

Note.   Since  one  factor  is  seen  in  (2)  to  be  c  +  d,  dividing  by  c  +  <? 
we  obtain  the  other  factor,  a  +  h. 


72  ALGEBRA. 

(2)  Find  the  factors  of  ac  -f-  ad  —he  —  bd. 

ac  +  ad  —  hc  —  hd—  (ac  +  ad)  —  (be  -f-  bd) 
=  a(c'^d)-b(c-^d) 
=  (a-b)(e-^d). 

Note.  The  signs  of  the  last  two  terms,  —  be  —  bd,  are  changed  to  +, 
when  put  within  a  parenthesis  preceded  by  the  minus  sign. 

(3)  Resolve  into  factors  bx^  —  lb ax"^  —  x -\- 3 a. 

b x^  —  lb ax^  —  X  -{-  3  a  =  (bx^  —  lb ax"^)  —  (x  ~  3 a) 
-=bx''{x-2>a)-l{x~Za) 
=  {bx'-l){x-2>a). 

(4)  Resolve  into  factors  %y  —  21  x^y  —  10 x  -\-  45a:l 

6?/ -  21x''y -  lOx  +  45^'  =  (45a;'  -  21x'y)-{10x - 6y) 
=  ^x\bx-2,y)-2{bx-2>y) 
=  (9a;^-2)(5a;-3y). 

Note.  By  grouping  the  terms  thus,  (Gy  —  27x'y)  —  (10  a;  -  45  a:'), 
we  obtain  for  the  factors  (Sy  —  5x)(2  —  9  a;'').  "^ 

But  (3  2/  -  5a;)(2  -  9a;2)  =  (Ox^  -  2)(5a;  -  3y),  since,  by  the  Law  of 
Signs,  the  signs  of  twofactorSy  or  of  any  even  7iumber  of  factors,  may 
be  changed  without  altering  the  value  of  the  product. 

Exercise  28. 

Resolve  into  factors : 

1 .  x'^  —  ax  —  hx-{-  ab.  6.  ahx  —  aby  -{-pqx  —  pqy. 

2.  ab-\-ay  —  by  —  y^  7.  cdx^  -f  adxy  —  bexy  —  a5y^ 

3.  be-\-bx  —  ex  —  o^.  8.  obey —  h^dy  —  acdx-\-bd}x. 

4.  mx ^ mn  ■\- ax -\- an.  9.  ax  — ay —  bx -[-by. 

5.  cc?a;^  —  c:cy+  dxy  —  y"^.  10.  cc?2;^  —  cyz  -f-  c?yz  —  y^ 


FACTORS.  73 

96.  Perfect  Squares.  If  an  expression  can  be  resolved 
into  two  equal  factors,  the  expression  is  called  a  perfect 
square,  and  one  of  its  equal  factors  is  called  its  square  root. 

Thus,  16:ry  —  4iX^y  X  4:r'y.  Hence,  16a:y  is  a  perfect 
square,  and  ^x^y  is  its  square  root. 

Note.  The  square  root  of  IQn^y"^  may  be  —  4a^y  as  well  as  +  40,"^^, 
for  —  4x'3/  X  -  4a^2/  =  IQa^y^ ;  but  throughout  this  chapter  the  posi- 
tive square  root  only  will  be  considered. 

97.  The  rule  for  extracting  the  square  root  of  a  perfect 
square,  when  the  square  is  a  monomial,  is  as  follows : 

Extract  the  square  root  of  the  coefficient,  and  divide  the 
index  of  each  letter  hy  2. 

98»  In  like  manner,  the  rule  for  extracting  the  cube 
root  of  a  perfect  cube,  when  the  cube  is  a  monomial,  is, 

Extract  the  cube  root  of  the  coefficient,  and  divide  the 
index  of  each  letter  hy  3. 

99.  By  .§§  79,  80,  a  trinomial  is  a  perfect  square,  if  its 
first  and  last  t^^ms  are  perfect  squares  and  positive,  and 
its  middle  term  is  twice  the  product  of  their  square  roots. 
Thus,  16  a'^  —  24a6  +  ^^^  is  a  perfect  square. 

The  rule  for  extracting  the  square  root  of  a  perfect 
square,  when  the  square  is  a  trinomial,  is  as  follows : 

Extract  the  square  roots  of  the  first  and  last  terms,  and 
connect  these  square  roots  hy  the  sign  of  the  middle  term. 

Thus,  if  we  wish  to  find  the  square  root  of 
16a2-24a5 -1-952, 
we  take  the  square  roots  of  16  a^  and  9  h^,  which  are  4  a  and  3  h, 
respectively,  and  connect  these  square  roots  by  the  sign  of  the  middle 
term,  which  is  — .     The  square  root  is  therefore 
4a-36. 

In  like  manner,  the  square  root  of 

16a2  +  24a6  +  9Z>2is4a  +  36. 


74  ALGEBRA. 

Case  III. 
100.  When  a  Trinomial  is  a  Perfect  Square. 

(1)  Resolve  into  factors  x^-{-2x7/-\-  yl 
From  §  99,  the  factors  of  x^  -{•2xy  -}-  y^  are 

{x^y){x-\-y). 

(2)  Resolve  into  factors  x^  —  2x^y  +  y^. 
From  §  99,  the  factors  oi  x^  —  2  x^y  +  ?/^  are 

{x'-y){x^-y). 

Exercise  29. 
Resolve  into  factors  : 

1.  a;' +  12a; +  36.  8.  y*  + 16yV  + 64^*. 

2.  a;' +  28^ +196.  9.  /  + 24y''  + 144. 

3.  a;^  +  34a;  +  289.  10.  a;V  +  162 0:2  +  6561. 

4.  2^  +  2^  +  1.  11.  4a'  +  12a5^  +  95*. 

5.  3/^ +  200y+ 10,000.  12.  ^ xhf  ■\- Z^ xyH  ^  2^ z\ 

6.  2* +  142' +  49.  13.  9a;2_|_i2a;y  +  4y\ 

7.  a;'  +  36a:y  +  324?/l  14.  4 aV  +  20 aVy  +  25 a:y . 

Exercise  30. 
Resolve  into  factors : 

1.  a'- 8a +16.  7.  ?/' -  50y2  + 6252^ 

2.  a^- 30a +  225.  8.  :r*  -  32a,y  +  256/. 

3.  a;' -38a; +  361.  9.    s«- 342^  +  289. 

4.  a;'  — 40 a; +400.  10.  4a;y  —  20a;y2  +  252/V. 

5.  2/^-100y  +  2500.  11.  16a;y- 8a7yV  +  yV. 

6.  y'  -  20y'  +  100.  12.  9 a'5V - 6  a5Vc^+  h'c'a  . 


FACTORS.  75 

13.  16a;«-8a;y  +  a:y.         16.    l~6ab'  +  9a'b\ 

14.  a^x*-2a'bxY  +  by.      17.    9mV  -  247?27i  +  16. 

15.  S6xy-60xy'  +  26y\    18.    4.b'x' -Ubx^i/ +  9xy. 

19.  49 a'- 112 ab-i-6U\ 

20.  64 .t'?/^  -  160 xYz  +  100 a;V. 

21.  49  a'bV- 28  abcx-i- 4  x\ 

22.  121  :i:*  -  286 a;^  +  169 y^ 

23.  289a;VV  -  102a;yVc^  +  9y'z'd\ 

24 .  361  .ry 2'  -  76  abcxi/z  +  4  a'6'cl 

Case  IV. 

101.  When  a  Binomial  is  the  Difference  of  Two  Squares. 
(1)  Resolve  into  factors  x^  —  yl 

From  §  81,  (a;  +  y)(x  —  y)  —  x^  —  y^. 

Hence,  the  difference  of  two  squares  is  the  product  of 
two  factors,  which  may  be  found  as  follows : 

Take  the  square  root  of  the  first  term  and  the  square  root 
of  the  second  term. 

The  sum  of  these  roots  will  form  the  first  factor. 

The  difference  of  these  roots  will  form  the  second  factor. 

102.  If  the  squares  are  compound  expressions,  the  same 
method  may  be  employed. 

(1)  Resolve  into  factors  {x  -\-  Sy)^  —  16  a'^ 

The  square  root  of  the  first  term  is  a;  +  3  y . 

The  square  root  of  the  second  term  is  4  a. 

The  sum  of  these  roots  is  a;  +  Sy  +  4a. 

The  difference  of  these  roots  is  a;  +  3y  —  4a. 

Therefore  {x  +  Zyf  -  IGa^  =  (x  +  3  y  +  4a)(a;  +  3y  -  4a). 


76  ALGEBRA. 

(2)  Kesolve  into  factors  a?  —  {?>b  —  bey. 

The  square  roots  of  the  terms  are  a  and  (36  —  5  c). 
The  sum  of  these  roots  is  a  +  (3  6  —  5  c),  or  a  +  3  6  —  5  c. 
The  difference  of  these  roots  is  a  —  (3  6  —  5  c),  or  a  —  3  6  +  5  c. 
Therefore  a?  -  (Zb-bcf  =  (a  +  36  -  5c)(a  -  36  +  5c). 

103.  If  the  factors  contain  like  terms,  these  terms  should 
be  collected  so  as  to  present  the  results  in  the  simplest 
form. 

(3)  Resolve  into  factors  (3  a  +  5  5)'^  -  (2  a  -  3  bf^ 
The  square  roots  of  the  terms  are  3  a  +  5  6  and  2  a  —  3  6. 
The  sum  of  these  roots  is  (3  a  +  5  6)  +  (2  a  —  3  6) 

or3a  +  56  +  2a  — 36  =  5a  +  26. 
The  difference  of  these  roots  is  (3  a  +  5  6)  —  (2a  —  3  6), 

or3a  +  56  —  2a  +  36  =  a  +  8  6. 
Therefore  (3a  +  56)^  -  (2a -  3 6)2  =  (5a  +  26)(a  +  86). 

104.  By  properly  grouping  the  terms,  compound  expres- 
sions may  often  be  written  as  the  difference  of  two  squares, 
and  the  factors  readily  found. 

(1)  Resolve  into  factors  a^  ~2ab  -\- b'^  —  9  c^. 

a^  -2ab  -\- b""  -  9 c^  =^  {a^  -  2ab  +  6^)  -  9c' 
=  (a -by -9c' 
=  (a-b  +  ScXa-b-Sc). 

(2)  Resolve  into  factors  12  ab -{- 9  x' —  4:  a' -  9b\ 
Note.  Here  12  a6  shows  that  it  is  the  middle  term  of  the  expres- 
sion which  has  in  its  first  and  last  terms  a?  and  6^,  and  the  minus 
sign  before  ^a?  and  9  6^  shows  that  these  terms  must  be  put  in  a 
parenthesis  with  the  minus  sign  before  it,  in  order  that  they  may  be 
made  positive. 

The  arrangement  will  be 

9x'-  (4a^  -  12ab  +  9b')^  9x''  -(2a-Sby 

=-  (Sx-j-2a-3bX3x-2a+Sb). 


FACTORS.  77 

(3)  Kesolve  into  factors  —  a^  -\-  b^  —  c^  -\-  d^  -{-  2ac  -{-  2  bd. 

Note.     Here  2  ac,  2  hd,  and  —  a^,  —  c"'',  indicate  the  arrangement 
required. 

=  (6'^  +  2bd+d')  -  (a'  -  2ac  +  c^) 

=  (^  +  0?  +  a  -  cXb  i-d-a  +  c). 

Exercise  31. 
Resolve  into  factors : 

1.  a'-b'.  18.  x'-2xy-{-2/-z\ 

2.  a' -16.  19.  a'  +  12bc-^b'~9c\ 

3.  4a' -25.  20.  a' -2ay  +  y'— a;'  — 2a:z-2l 

4.  a* -J*.  21.  2a;y-a;'-y'  +  2l 

5.  a'-l.  22.  :r'  +  3/'-2'-(^'-2a;y-2c/2. 

6.  a'- 61  23.  x''-f-\-z''-a''-2xz-]-2ay. 

7.  a«-l.  24.  2ab  +  a''-\-b'-c\ 

8.  36a;^- 492/1  25.  2a;y-a:'-2/'  +  a'^  +  6'-2a6. 

9.  100a;y-121a'6l  26.  (aa; +  %)'-!. 

10.  l-49a;l  27.  l-:r''-2/'  +  2a;y. 

11.  a* -25 61  28.  (5a-2y-(a-^)\ 

12.  (a-6)2-cl  29.  d'  —  2ab  +  b'~x\ 

13.  a;' -(a -6)1  30.  (a;  +  1)' -  (y  +  1)'. 

14.  (a-{-by~(c  +  dy.  31.  (a;  +  1)'^  -  (y  -  1)1 

15.  (x  +  yy-(x-yy.  32.  c^'- a:^  + 4:^2/- 4y^ 

16.  2a6-a'-6'  +  l.  33.  a"  -  b' -  2bc  -  c\ 

17.  x^~2yz-f-z^.  34.  4a;* -9a;' +  6a; -  1. 


78  ALGEBRA. 


Case  V. 


105.  When  a  Trinomial  has  the  Porm  a'  +  a'V  +  V.  A 
trinomial  in  the  form  of  a*  +  d^h^  +  ^*  can  be  written  as 
the  difference  of  two  squares. 

Since  a  trinomial  is  a  perfect  square  when  the  middle 
term  is  twice  the  product  of  the  square  roots  of  the  first 
and  last  terms,  it  is  obvious  that  we  must  add  d^h^  to  the 
middle  term  of  a*  +  o?})^  +  h^  to  make  it  a  perfect  square. 
We  must  also  subtract  a^h^  to  keep  the  value  of  the 
expression  unchanged.     We  shall  then  have 

(1)  a*  +  db'  -\.h'  =  a'  +  2dh^  +  5* -  a'h' 

=  (d  +  by  -  d'b' 

=  {d  +  ^>'  +  ab)  (d  +  b'-  ab) 
^{a'^ab-^b'){a'-ab-\-b'). 

If  in  the  above  expression  we  put  1  for  b,  we  shall  have 

(2)  •    a*4-a'  +  l  =  K  +  2a"'+l)-a'  ♦ 

=  {a'-^iy-d' 
-(a^4-l4-a)(a^  +  l-a) 
=  {d-\-a-^\){d'-a-\- 1). 

(3)  Resolve  into  factors  4  6*  -  37  b'c'  +  9  c\ 

Twice  the  product  of  the  square  roots  of  46*  and  9  c*  is  12  6V. 
We  may  separate  the  term  —376V  into  two  terms,  —12  6V  and 
—  25  6^c^,  and  write  the  expression 

(45*-126V  +  9c*)-25^>V 

=  (26'-3c7-25^'V 

=  {2h^  -  Z(^  +  bbc)  (2b'  -Sc'-  bbc) 

^(2b'  +  5bc-Sc')(2b'-6bc-Scy 


FACTORS.  79 

Exercise  32. 
Resolve  into  factors : 

1.  x'^xY-\-y\  8.  49m*+110mV+81n*. 

2.  9a;*  +  3a;y  +  4y*.  9.  9a*  + 21aV  + 25c^ 

3.  16a;*-17:cy4-2/*.  10.  49a*  -  ISa^^^^  12U*. 

4.  81a*  +  23a'^Z>^  +  16i*.  11.  64a:*  + 128a;y +  8I3/*. 

5.  81a*-28a'Z>»4-165*.  12.  4:r*- 37a:y  +  9?/*. 

6.  9a;*+38a;y  +  49/.  13.  25:r*  -  4l2;y  +  16?/*. 

7.  25a*-9a'62+ig^4^  ^^  81a:*- 34a:y +  /. 

Case  VI. 

106.  When  a  Trinomial  lias  the  Porm  x^  4-  ax  +  b. 

From  §  84  it  is  seen  that  a  trinomial  is  often  the  product 
of  two  binomials.  Conversely,  a  trinomial  may,  in  certain 
cases,  be  resolved  into  two  binomial  factors. 

107.  If  a  trinomial  of  the  form  0^ -\-ax-\-h  is  such  an 
expression  that  it  can  be  resolved  into  two  binomial  fac- 
tors, it  is  obvious  that  the  first  term  of  each  factor  will  be 
X,  and  that  the  second  terms  of  the  factors  will  be  two 
numbers  whose  product  is  h,  the  last  term  of  the  trinomial, 
and  whose  algebraic  sum  is  a,  the  coefficient  of  x  in  the 
middle  term  of  the  trinomial.    . 

(1)  Resolve  into  factors  a^  +  lla  +  30. 

We  are  required  to  find  two  numbers  whose  product  is  30  and 
whose  sum  is  11. 

Two  numbers  whose  product  is  30  are  1  and  30,  2  and  15,  3  and 
10,  5  and  6  ;  and  the  sum  of  the  last  two  numbers  is  11.     Hence, 

a^  +  11a  +  30  ■=  (a  +  5)(a  -f  6). 


80  ALGEBRA. 

(2)  Resolve  into  factors  x"^  —  7x-\-  12. 

"We  are  required  to  find  two  numbers  whose  product  is  12  and 
whose  algebraic  sum  is  —  7. 

Since  the  product  is  +  12,  the  two  numbers  are  both  positive  or  both 
negative ;  and  since  their  sum  is  —  7,  they  must  both  be  negative. 

Two  negative  numbers  whose  product  is  12  are  —  12  and  —1,-6 
and  —2,-4  and  -  3  ;  and  the  sum  of  the  last  two  numbers  is  —  7. 
Hence, 

x''  -1 X  +12  =  {x  -  ^){x  -  2>). 

(3)  Resolve  into  factors  x^  --{-^x  —  24. 

"We  are  required  to  find  two  numbers  whose  product  is  —  24  and 
whose  algebraic  sum  is  2. 

Since  the  product  is  —  24,  one  of  the  numbers  is  positive  and  the 
other  negative ;  and  since  their  sum  is  +  2,  the  larger  number  is 
positive. 

Two  numbers  whose  product  is  —  24,  and  the  larger  number  posi- 
tive, are  24  and  —  1,  12  and  —  2,  8  and  —  3,6  and  —  4 ;  and  the  sum 
of  the  last  two  numbers  is  +  2.     Hence, 

a?  +  2x-24:={x  +  e>){x  -  4). 

(4)  Resolve  into  factors  a;^  —  3  a;  —  18. 

We  are  required  to  find  two  numbers  whose  product  is  —  18  and 
whose  algebraic  sum  is  —  3. 

Since  the  product  is  —  18,  one  of  the  numbers  is  positive  and  the 
otlier  negative ;  and  since  their  sum  is  —  3,  the  larger  number  is 
negative. 

Two  numbers  whose  product  is  —  18,  and  the  larger  number  nega- 
tive, are  — 18  and  1,-9  and  2,  —  6  and  3  ;  and  the  sum  of  the  last 
two  numbers  is  —  3.     Hence, 

x^-2>x-l^  =  {x-  6)(a;  +  3). 

Therefore  in  general, 

a^  +  {a-{-h)x  -\-  ah  =  {x  -]-  a){x  -{-h) 

whatever  the  values  oi  a  and  h. 


FACTORS.  81 

Exercise  33. 

Resolve  into  factors : 

1.  :^^  +  ll.r4-24.  14.    a'+ba'  +  G. 

2.  a;^+llrr  +  30.  15.    z^  +  Az'  +  S. 

3.  y'+l1y-{-60.  16.    a'b' +  18 ab -]- S2. 

4.  2^+13z  +  12.  17.    a;y  +  7a;y  +  12. 

5.  0;^  + 21a; +  110.  18.    2^° +  102^+16. 

6.  y'^^  352/ +  300.  19.   a' +  9  ab +  20 b\ 

7.  6^  +  235  +  102.  20.    a;«  +  9a;'  +  20. 

8.  a;'  +  3a:  +  2.  21.    aV  + 14a5a;  +  33^>l 

9.  a:'  +  7ir  +  6.  22.    aV  +  7aca;+ 10a;l 

10.  a'-}-9ab-{-8b\  23.    rry2'^  + 19a;yz  +  48. 

11.  a;^  +  13aar  +  36a^  24.    ^>V  + 18a5c  +  65al 

12.  ?/'  + 19^32/ +  48/.  25.    ?V  +  23m  +  902l 

13.  2^  +  29^^  +  100^^  26.    mV+20mVp^+51py. 

Exercise  34. 

Resolve  into  factors  : 

1.  a:'-7:r+10.  7.    a:*-4aV  +  3a*. 

2.  07^-293; +  190.  8.    a;'^-8a:  +  12. 

3.  a^- 23a +  132.  9.    z^-57z-\-66. 

4.  6^-305  +  200.  10.   2/«-7y'  +  12. 

5.  2^-43z  +  460.  11.    a:y-27a:y  +  26. 

6.  x'-7x-i-6.  12.    a*b'-lla'b'-\-SO. 


8^  ALGEBRA. 

13.  a^b'c' -  IS  abc  + 22.  19.    a;'-20ar+91. 

14.  x'-16x  +  50.  20.    ri;'- 23 a; +120. 

15.  o;'^  — 20:^+100.  21.    2^-532  +  360. 

16.  aV-21aa;  +  54.  22.    x""  -  (a-j- c)x-^  ac. 

17.  aV  -  IQabx  +  39  5'^  23.    ^/V  -  28a5y2  +  187 a'^'^ 

18.  aV-24ac2+1432^  24.    c^c/^  -  30a^>c^+ 221a^5^ 

Exercise  35. 
Resolve  into  factors : 

1.  x'-^Qx-l.  8.    a'  +  25a-150. 

2.  a;' +  5a: -84.  9.    b^-\-Sb*-4:. 

3.  3/^ +72/ -60.  10.    ^V  + 3  5c -154. 

4.  2/2  +  12^-45.  11.    0^*^+15^^-100. 

5.  2^+112-12.  12.    0^^  +  17(7-390. 

6.  2^  +  132-140.  13.    aHa-132. 

7.  a*^+13a-300.  14.   a;VV  + 9:ry2  -  22. 

Exercise  36. 
Resolve  into  factors : 

1.  a;'^-3a:-28.  6.    a'^- 15a -100. 

2.  2/'-7y-18.  7.    c''~9c'-'l0. 

3.  a;2-9:r-36.  8.    a;^-8a;-20. 

4.  2^-112-60.  9.   y'-6a^-b0a\ 
6.    2^-132-14.  10..    a''5'^-3a^> -4. 


FACTORS.  83 

11.  aV-3aa:-54.  14.   2/V-53/V-84. 

12.  c'd'-24:cd-180.  15.   a'b'-ieab-Se. 

13.  aV-a'c-2.  16.    a;'- (a  —  &)a;  -  a^. 


Case  VII. 
108.  When  a  Trinomial  has  the  Perm  ax^  +  bx  +  c. 

From  §  85, 

(Sx  -2)(bx  +  3) 

=  Ibx'  -i-9x-10x-6=15x'-x-6.     (1) 

(3a; -2)  (5a; -3) 

=  15a:2_9a:-10a;+6  =  15a:^-19a;  +  6.  (2) 

Consider  the  resulting  trinomials  : 

The  first  term  in  (1)  and  (2)  is  the  product  3  a;  X  5  a;. 

The  middle  term  in  (1)  is  the  algebraic  sum  of  the  products 

3a;  X  3  and  (-2)x5a;. 

The  middle  term  in  (2)  is  the  algebraic  sum  of  the  products 

3a;X(-3)and(-2)x5x.  , 

The  last  term  in  (1)  is  the  product  (—  2)  x  3. 
The  last  term  in  (2)  is  the  product  (-  2)  x  (-  3). 
The  trinomials  have  no  monomial  factor,  since  no  one  of  their 
factors  has  a  monomial  factor.     Hence, 

1.  If  the  third  term  of  a  given  trinomial  is  negative, 
the  second  terms  of  its  binomial  factors  will  have  unlike 
signs. 

2.  If  the  third  term  is  positive,  the  secoyid  terms  of  its 
binomial  factors  will  have  the  same  sign,  and  this  sign  is 
the  sign  of  the  middle  term. 

3.  If  a  trinomial  has  no  monomial  factor,  neither  of  its 
binomial  factors  can  have  a  monomial  factor. 


84  ALGEBRA. 

(1)  Resolve  into  factors  6a;^+  17:r-l-  12. 

The  first  terms  of  the  binomial  factors  must  be  either  6x  and  x,  or 
3  X  and  2  x. 

The  second  terms  of  the  binomial  factors  must  be  12  and  1,  or  6 
and  2,  or  3  and  4. 

We  therefore  write 

I.   (6x+      )(x+     ),     or     11.   (Sx+     ){2x+     ). 

For  the  second  terms  of  these  factors  we  must  reject  1  and  12 ;  for 
12  put  in  the  second  factor  of  I.  would  make  the  product  6  a;  X  12  too 
large,  and  put  in  the  first  factor  of  I.,  or  in  either  factor  of  II.,  the 
result  would  show  a  monomial  factor. 

We  must  also  reject  6  and  2 ;  for  if  put  in  I.  or  II.  the  results 
would  show  monomial  factors  ;  and  for  the  same  reason  we  must 
reject  3  and  4  for  I. 

The  required  factors,  therefore,  are  (3  a;  +  4)  and  (2x  -f  3). 

(2)  Resolve  into  factors  14a;^  —  llo;  —  15. 

For  a  first  trial  we  write 

(7  a;    )(2a;    ). 

Since  the  third  term  of  the  given  trinomial  is  —  15,  the  second 
terms  of  the  binomial  factors  will  have  unlike  signs,  and  the  two 
products  which  together  form  the  middle  term  will  be  one  +,  and 
the  other  — .  Also,  since  the  middle  term  is  — 11  a;,  the  negative 
product  will  exceed  in  absolute  value  the  positive  product  by  —  11  a;. 

The  required  factors,  therefore,  are  (7  a;  +  5)  and  (2  a;  —  3). 

Exercise  37. 
Resolve  into  factors : 

1.  12a;'- 5a:  — 2.  6.  Q>  x' -{- 5 x  -  A. 

2.  12a;'^-7a:  +  l.  7.  4:x'+lSx  +  S. 

3.  Ux'-x-l.  8.  4:x^+llx~3. 

4.  3a:' -2a: -5.  9.  4a:' -4a:  — 3. 

5.  3a:' +  4a: -4.  10.  a:' -  3 aa:  +  2 a'. 


11. 

12a*  +  «V-a;*. 

12. 

2a,'2  +  5:iy4-2?/'. 

13. 

6aV+aa;  — 1. 

14. 

W-1hx-2>x\ 

15. 

4x^4- 8a: +  3. 

16. 

o?-ax  —  Q>x\ 

17. 

d>a'^Uah~lbh\ 

FACTORS.  .    85 

18.  6a2 -19ac+ 10  cl 

19.  8a;2  +  34a;3/  +  2l3/^ 

20.  8a;'-22a:y-21y^ 

21.  ^x^+l^xy-ly"^. 

22.  lla2-23a^>^-26^ 

23.  2c'^-13cc^^-6(i^ 

24.  Qf-\-1yz-^z\ 

Case  VIIL 
109.   When  a  Binomial  is  the  Sum  or  Difference  of  Two  Cubes. 

From  §88,     ^.±^ -=  a' -  ah -{- h' ; 
a-\-h 

and  ^^SzR=a'j^ah  +  h\ 

a  —  h 

.\  a'-\-h'^{a-\-h){a:'-ah-\-h'') 

and  o?-h^  =  {a-  bXa'  -{-ab  +  b'). 

In  like  manner  we  can  resolve  into  factors  any  expres- 
sion which  can  be  written  as  the  sum  or  the  difference  of 
two  cubes. 

(1)   Resolve  into  factors  8  aH  27  6«. 

Since  by  g  98,  8a^=={2af,  and  27  b^  =  {3  b'^f,  we  can  write 
8a'  +  276«a8(2a)3  +  (3  62)3. 

Since  a^  +  b^  =  {a  +  b){a^  -ab  +  b% 

we  have,  by  putting  2  a  for  a  and  3  b^  for  b, 

{2af  +  (3  62)3  =  (2  a  ^  3  b^){i  a'-Gab^  +  d  6*). 


86  ALGEBRA. 

(2)  Eesolve  into  factors  12^  x^  —  1. 
125a!3-l  =  (5a;)3-l 

=  (5aj-l)(25x2  +  5a;  +  l). 

(3)  Resolve  into  factors  x^  +  y^. 

=  (x2  +  3/3)(a;*  -  x'^y^  +  y^). 

110.   The  same  method  is  applicable  when  the  cubes  are 
compound  expressions. 

(4)  Resolve  into  factors  {x  —  yf  +  z^. 

Since    ■o?  +  W  =  {a  +  h){a?  -ab  +  b-"), 
we  have,  by  putting  x  —  y  for  a  and  z  for  b, 

{x-yf  +  ^  =  [{x  -y)+  z][{x  -  y)^  -  {x  -  y)z  +  z^] 

=  {x  —  y-\-  z){x^  —  2xy+y^  —  xz+yz  +  z^). 

Exercise  38. 
Resolve  into  factors  : 

1.  rc'  +  S.  8.  27  a' -1728. 

2.  rr'  +  216.  9.  21oJ'-h\ 

3.  2/' +  64^1  10.  {x-\-yY-l. 

4.  646^+125^1  11.  (a;  +  yy  +  l. 

5.  S:i?-21f.  12.  ^o^-{a-h)\ 

6.  64y'-10002l  13.  {x-^yy-\-c\ 

7.  729a:' -512y^  14.  (5;  +  3/y-(rr-y)3. 


FACTORS.  87 

Case  IX. 

111.  When  a  Polynomial  is  the  Product  of  Two  Trinomials. 
The  following  method  is  convenient  for  resolving  a  poly- 
nomial into  its  trinomial  factors : 

Find  the  factors  of 

2a;^  -  ^xy  +  23/^  +  7:rz  -  byz  -f  Zz\ 

1.  Reject  the  terms  that  contain  z. 

2.  Reject  the  terms  that  contain  y. 

3.  Reject  the  terms  that  contain  x. 

Factor  the  expression  that  remains  in  each  case. 

1.  2x^-5xy +  2y^  =  {x-2y){2x-y). 

2.  2x'^  +  7xz  +Sz^  ={x  +  3z){2x  +  z). 

3.  2?/2 - 5yz  +Sz^  =l2y-3z){y- z). 

Arrange  these  three  pairs  of  factors  in  two  rows  of  three  factors 
each,  so  that  any  two  factors  of  each  row  may  have  a  common  term 
including  the  sign. 

Thus,  x-2y,  x  +  Sz,  ~2y +Sz; 

2x  —  y,  2x  +  z,  ~y  +  z. 

From  the  first  row,  select  the  terms  common  to  two  factors  for 
one  trinomial  factor : 

x-2y  +  32. 
From  the  second  row,  select  the  terms  common  to  two  factors  for 
the  other  trinomial  factor  : 

2x  —  y  +  z. 

Hence,  2x^  —  5xy  +  2y^  +  ^a^z  —  5yz  -{-  Sz^ 

=  (S;-2y  +  Sz)(2x-y  +  z). 

112.  When  a  factor  obtained  from  the  first  three  terms 
is  also  a  factor  of  the  remaining  terms,  the  expression  is 
easily  resolved. .   Thus, 

x'^-3xy  +  2y^-3x  +  Qy 

=  {x-2y){x-y)-S{x-2y) 

=  {x-2y){x-y-S).  '     , 


88  ALGEBRA. 

Exercise  39. 
Resolve  into  factors  : 

1.  2a;^-5;r?/  +  2y2-17:i;+13y  +  21. 

2.  6a;'-37rcy  +  63/'-5a;-5y-l. 

3.  ^x^  —  bxy  —  Q)y^  —  x~by  —  \. 

4.  5a;'-8:ry  +  3yH7:r-5y  +  2. 

6.  2x''-xy-Zy''-d>x-\-ly-\-^. 
vl  6.  x' -2by^ -lOx -20y ~^2l. 

7.  2x^  —  5a;?/  4-'2y^  —  3:2;  —  ys;  —  z^. 

8.  6a;'^-a;y-y-3:r2  +  6y^-92^ 

9.  ^x^-1xy  +  y'-^2>bxz~byz-e>z\ 
10.-  5a;'-8a;y  +  3?/'~  3^2  +  y2-22l 

11.  2a;'-a;y-3y'-5y2-2;2^ 

12.  6a;*^-13.ry  +  6y'+12a;2-13y2  +  62^ 

13.  a;'  — 2a;y 4-2/^  +  5a;  — 5?/. 

14.  2x^-\-bxy-^y-'-4.xz^2yz. 

Case  X. 

113.   Binomials  of  the  Form  x**  —  y"  or  x'*  -f-  y**,  and  n  >  3. 

1.  When  a  binomial  has  the  form  x^  —  y**,  but  cannot  be 
written  as  the  difference  of  two  perfect  squares,  or  of  two 
perfect  cubes,  it  is  still  possible  to  resolve  it^  into  two  fac- 
tors, one  of  which  is  a;  — y.     Thus  (§  89), 


FACTORS.  89 

2.  When  a  binomial  has  the  form  x""  +  3/**,  but  cannot  be 
written  as  the  sum  of  two  perfect  cubes,  it  is  still  possible 
to  resolve  it  into  two  factors,  except  when  n  is  2,  4,  8,  16, 
or  some  other  power  of  2.     Thus  (§  89), 

a'-^b'  =  (a  +  b){a'  -  a^h  +  a'b'  -  ah^  +  h').  ■ 

But  o?-\-h'^,  a*  +  &*,  a^-\-}f,  cannot  be  resolved  into 
factors. 

Note  1.  The  student  must  be  careful  to  select  the  best  method 
of  resolving  an  expression  into  factors.  Thus,  a^  —  h^  can  be  written 
as  the  difference  of  two  squares,  or  as  the  difference  of  two  cubes,  or 
be  divided  by  a  —  6,  or  by  a  +  h.  Of  all  these  methods,  the  best  is  to 
write  the  expression  as  the  difference  of  two  squares,  as  follows  : 

(a3)2  -  (63)2  _  ((j3  +  53)(^3  _  ^3) 

=  (a  +  h){o?  -ab  +  ¥){a  -  h){a?  +  ah  +  ¥). 

Note  2.  From  the  last  example,  it  will  be  seen  that  an  expres- 
sion can  sometimes  be  resolved  into  three  or  more  factors. 

x-8-68=(a;*  +  6*)(a;*-6*) 

=  (x*  +  6*)(a;2  +  62)(a;2  -  6^) 

=  (a;*  +  h^x"  +  62)(x  +  h){x  -  h). 

Note  3.  When  a  factor  occurs  in  every  term  of  an  expression, 
this  factor  should  first  be  removed.     Thus, 

8  a;2  -  50a2  +  4a;  -  10a  =  2(4x2  _  25a2  +  2.r  -  5a) 

=  2 [(4aj2  -  25 a2)  +  (2a; -  5 a)] 
=  2(2a;-5a)(2a;  +  5a  +  l). 

Note  4.  Sometimes  an  expression  can  be  easily  resolved  if  we 
replace  the  last  term  but  one  by  two  terms,  one  of  which  shall  have 
for  a  coefficient  an  exact  divisor  or  a  multiple  of  the  last  term.    Thus, 


90  ALGEBRA. 

(1)  a^  -  dx""  +  llx  -15  =  {a^  -  dx-"  +  6x)  +  {5x  -  15) 

=  x(x'^-5x  +  6)  +  5{x-3) 
=  {x-S)[x{x-2)  +  b] 
=  {x-3){x'''-2x  +  5). 

(2)  a^-26x-5  =  {a^-25x)-{x  +  5) 

=  x{x^-2b)-{x  +  5) 
=  {x  +  5){x^-5x-l). 

(3)  a3  +  3a;2  -  4  =  (x3  +  2x2)  +  (^2  _  4) 

=  x''{x  +  2)  +  {x''-4:) 
=  {x  +  2)(a;2  +  a;  -  2) 
=  {x  +  2)(x  +  2){x-l). 


Exercise  40. 

MISCELLANEOUS   EXAMPLES. 

Find  the  factors  of 

1.  5x'-lbx-20.            ^  9.  a'  +  a'  +  l. 

2.  2x''-16x'  +  24:x\  10.  x'-y'-xz  +  yz. 

3.  Sa''b''-9ab-12.  11.  ab~ac-b'  +  bc. 

4.  a'  +  2a:r  +  a;'  +  4a  +  4a;.  12.  Sx"" —  Sxz  —  xy -}-yz. 

5.  a^  — 2a5  +  ^>'^-c^  13.  a"  —  x' -  ab  -  bx. 

6.  a;'-2a;y-f2/•''-c''^-2co?~c^^  14.  a' -  2aa;  +  a:H«-^. 

7.  4:-x''-2x^-x\  15.  3a;'-33/'-2:r  +  2y. 

8.  a'-b^-a-b.  16.  a;*  +  a;' +  a;' +  a;. 

)  17 .   a*x^  —  aV  —  a'^a;''  +  1. 

18.   3a;'  -  2a:V  -  21  xy'  +  18yl 


FACTORS.                                                  91 

19. 

Ax'-x'  +  2x- 

-1. 

30. 

36aVy^-255V2/l 

20. 

x'-y\ 

31. 

9xy-S0xy'z-{-2bz\ 

21. 

x'  +  f. 

32. 

16a^-x. 

22. 

12^ -x\ 

33. 

x^-2  xy-2  xz+y'^2  yz-\-z\ 

23. 

x''y-\-y'\ 

34. 

a'-ah-U'~^a-^l2b. 

24. 

a'c-cK 

35. 

x^-\-2xy-^y''  —  x  —  y  —  ^. 

25. 

X'-^4:X-21. 

36. 

{a^hy-o\ 

26. 

3a'-21a6  +  305\ 

37. 

x''-xy~Qf-4:X-\-\2y. 

27. 

2x'-^x^y-QxY. 

38. 

\-x-\-x'-x\ 

28. 

^a'-iab  +  b\ 

39. 

Zx'-llxy-{-^y\ 

29.    16a;''- 80  a;?/ +  100 y^   40.    a;' +  20a; +  91. 
41.    {x  —  y){x'-z'')-{x  —  z){x^-y''). 

42.  a;'  — 5a;  — 24.  50.    y''  —  ^y-ll1. 

43.  {x'-y''-z^y-^yh\    51.    a;^  +  6a;-135. 

44.  5a;y  +  5a;y-60a;2^   52.   4a'*  -  12a^>  +  95' -4c'' 

45.  3a;'-a;''  +  3a;-l.         53.    {a-\-^hy -9{h  -  c)\ 

46.  x^~2mx^m^-n\      54.    "^ x^  —  ^y^ ■{- ^yz  —  z\ 

47.  ^a'b''-{p?-{-h''-cy.    55.    65V  -  75a;' -  3a;^ 

48.  a^  +  a^  56.    a? —  h^ -Zab{a-h). 

49.  l-14a'a;  +  49aV.       57.   a;'  +  y' +  3a;y(a; +  y). 

58.    a'  -b'  -  a{a'  -~h')  ^h{a  -h)\ 
59.    9a;y-3a;y'-6y*.        60.    ^x' -\-Uxy -\-Qy\ 

61.  6a'5'-a5'-125*. 

62.  o?  ■\-2ad+  €1""  ~  W  ^I2be  -9c\ 
63.    a;'-2a;V+4a;y'-8y^    64.    4aV  -  8a5a;  +  35^ 


92  ALGEBRA. 

65.  l^x^~2^xy-\-^y'^'^x-Qy.  74.  \^a'x-2x\ 

66.  2x'-\-2xy-l2y''-^^xz^lSyz.  75.  ^2bx^  -  Uy\ 

67.  {x-\-yy-l-xy{x-{-y-\-V).  76.  x-21x\ 
V  68.  x'  —  y''-z^-\-2yz-]-x-\-y-z.  77.  x^^  -  y'\ 

69.  2a;'+4a;y  +  22/^  +  2aa;+2a?/.  78.  49m'- 121  nl 

70.  16a'5  +  32a6c  +  125cl  79.  16-81/. 

71.  m^p  —  m'^q  —  n'^p-[-'n}q.  80.  122*  — 2'  — 6. 

72.  12aa;'— 14a.r?/-6ayl  81.  a;' —  a;' +  a; -  1. 

73.  2x^-\-A:X^-l(dx.  82.  a;'  +  2:c4-l— yl 

83.  49  {a  -  by  -  64(m  -  nf. 

84.  4(a^'  +  cc/)'-(a'4-52-c'-c^7. 

85.  a;' -53a; +  360. 

86.  x^-2x'y-^x'-4:X  +  Sy-4:, 

87.  2a5-2^>c-ae  +  ce  +  2^)^-5e. 

88.  l25x^  +  Sb0xy  +  24:bxy\ 

89.  a^  +  a'b  +  a*^>'  +  a'5^  +  a'b'  +  a6^ 

90.  2a\v  —  2a'cx  +  2ac'x-2c*x. 

91.  6:r'-5a;y-62/'  +  3a:2  +  15y2~92l 

92.  4a;'^-9a;y  + 22/^^-3:^2-2/2-22. 

93.  Sa''-1abi-2b'  +  bac-bbc  +  2c\ 

94.  a;*-2;r'  +  a7'-8a;  +  8. 

95.  5x'-8xy  +  S7/  —  bx-\-Sy. 

96.  tt'-2ac?+c^'-4^>H125c-9c^ 

97.  (a;'^  -  5;  -  6)(a;' -  a;  -  20). 


CHAPTER  VIII. 
COMMON   FACTORS  AND  MULTIPLES. 

114.  Oommon  Factors.  A  common  factor  of  two  or  more 
numbers  is  an  integral  number  which  divides  each  of  them 
without  a  remainder. 

115.  A  common  factor  of  two  or  more  expressions  is 
an  integral  and  rational  expression  which  divides  each  of 
them  without  a  remainder.  Thus,  5  a  is  a  common  fac- 
tor of  20  a  and  25  a;  3xy  is  a  common  factor  of  12  oc^y^ 
and  15a;y. 

116.  Two  numbers  are  said  to  be  prime  to  each  other 
when  they  have  no  common  factor  except  1. 

117.  Two  expressions  are  said  to  be  prime  to  each  other 
when  they  have  no  common  factor  except  1. 

118.  The  highest  common  factor  of  two  or  more  numbers  is 
the  greatest  number  that  will  divide  each  of  them  without 
a  remainder. 

119.  The  highest  common  factor  of  two  or  more  expres- 
sions is  the  expression  of  highest  degree  that  will  divide 
each  of  them  without  a  remainder.  Thus,  3  a^  is  the  highest 
common  factor  of  3a^  6a^  and  12  a*;  5a;y  is  the  highest 
common  factor  of  10 ry  and  15a;y. 

For  brevity,  we  use  H.  C.  F.  to  stand  for  "  highest  com- 
mon factor." 


94  ALGEBRA. 

To  find  the  highest  common  factor  of  two  algelDraic 
expressions : 

Case  I. 

120.  When  the  Pactors  can  be  found  by  Inspection. 

(1)  Find  the  H.  C.  F.  of  ^2a'b'  and  60a'b\ 

4:2a'b'  =  2xZx7xaaaXbb; 
eOa'b*  =  2x2x3x5XaaX  bbbb. 
.'.  the  H.  0.  F.  =  2  X  3  X  aa  X  bb,  or  6a'b\ 

(2)  Find  the  H.  C.  F.  of  2  a'x  +  2  ax'  and  3  abxT/  -f  3  bxY 

2a'x-\-2ax'^       =2ax{a-\-x); 
3  abxy  +  3  bx^y  —  3  bxy  {a-\-x). 
:.  theH.C.F.        =x{a-\-x). 

(3)  Find  the  H.  C.  F.  of  4a;^  +  4a;- 48,  6a;^- 48a;-f90. 

^x'-\-^x-^%-=^{x?^rX-\2) 

=  4(a;-3)(a;  +  4); 

6a;'  -  48a;  +  90  =  6 {x"  -  8a;  +  15) 
=  6(a;-3)(a;-5). 
.-.  theH.C.F.       =2 (a; -3) 
=  2a;-6. 

Hence,  to  find  the  H.  C.  F.  of  two  or  more  expressions : 

Resolve  each  expression  into  its  simplest  factors. 

Find  the  product  of  all  the  common  factors,  talcing  each 
factor  the  least  number  of  times  it  occurs  in  any  of  the  given 
expressions. 


COMMON   FACTORS   AND   MULTIPLES.  95 

Exercise  41. 
Find  the  H.  C.  F.  of 

1.  18 abVd  and  SQa''bcd\ 

2.  I7pq\  Mp'q,  and  51//. 

3.  8xyz\  12xYz\  and  20xyz\ 

4.  30:ry,  90a;y,  and  120a;y. 

5.  a'-b'  and  ft' -  51  7.    a'  +  a;'  and  (a  +  a^)'- 

6.  a'-x'  and  (a-rr^.        8.    9a;^-l  and  (3.1;  4-1)'- 

9.  7x''  —  4:x  and  7a^x-4:a\ 

10.  12aV2/-4aV  and  SOaV^/^- lOaVy'. 

11.  8a'b'c- 12 a'bc'  and  6a5*c  +  4a5V. 

12.  x''-2x  —  B  and  .'r'  +  a;~12. 

13.  2a'-2a5'^  and  U^a  +  bf. 

14.  12n'V(^-2/)(^-32/)  and  18x\x-y){Sx -y). 

15.  3a;='  +  6a;^-24a;  and  6a;'-96a:. 

16.  ac(a  —  b){a  —  c)  and  bc{b  —  a){b  —  c). 

17.  lOa;^ - 60a:y  +  5a;2/'  and  5:ry -^xf- 100/. 

18.  :r(a;+iy,  a;^(a;^-l),  and  2x(x^-x-2). 

19.  3a;^  -  6a:  +  3,  6a:^  +  6a;  -  12,  and  12a;^  -  12. 

20.  6  (a  -  b)\  8 (a'  -  b'f,  and  10 (a*  -  b'). 

21.  a;^-2/',  (^  +  2/)'.  and  x'  +  Sxy  +  2y\ 

22.  a;'  —  2/^  x^  —  2/^  and  a;^  —  Tary  +  Qy^. 

23.  a;'-l,  ir'-l,  and  x^  +  x-2. 


96  ALGEBRA. 


Case  II. 


121.  When  the  Factors  cannot  be  found  by  Inspection. 

The  method  to  be  employed  in  this  case  is  similar  to 
that  of  the  corresponding  case  in  Arithmetic.  And  as  in 
Arithmetic,  pairs  of  continually  decreasing  numbers  are 
obtained,  which  contain  as  a  factor  the  H.  C.  F.  required, 
so  in  Algebra,  pairs  of  expressions  of  continually  decreas- 
ing degrees  are  obtained,  which  contain  as  a  factor  the 
H.  C.  F.  required. 

122.  The  method  depends  upon  the  following  principles : 

(1)  Any  factor  of  an  expression  is  a  factor  also  of  any 
multiple  of  that  expression. 

Thus,  if  c  is  contained  3  times  in  A,  then  c  is  contained 
9  times  in  3^,  and  m  times  in  mA. 

(2)  Any  common  factor  of  two  expressions  is  a  factor  of 
their  sum,  their  difference,  and  of  the  sum  or  difference 
of  any  7nultiples  of  the  expressions. 

Thus,  if  c  is  contained  5  times  in  A,  and  3  times  in  B, 
then  c  is  contained  8  times  in  A-\-B,  and  2  times  in 
A~B. 

Also,  in  bA  +  2B  it  is  contained  5x5  +  2x3,  or  31  times, 
and  in  5  J.  —  2 -B  it  is  contained  5  X  5  —  2  X  3,  or  19  times. 

(3)  The  a.  C.  F.  of  two  expressions  is  not  changed  if  one 
of  the  expressions  is  divided  hy  a  factor  that  is  not  a  factor 
of  the  other  expression,  or  if  one  is  multiplied  by  a  factor 
that  is  not  a  factor  of  the  other  expression. 

Thus,  the  H.  C.  F.  of  4a^^c^  and  aVc?  is  not  changed  if 
we  remove  the  factors  4  and  b  from  4a^3c^  and  d  from 
a'^c^d ;  or  if  we  multiply  ^.c^bc^  by  7,  and  aVc?  by  11. 


COMMON   FACTORS   AND    MULTIPLES.  97 

123.  We  will  first  find  the  greatest  common  factor  of 
two  arithmetical  numbers,  and  then  show  that  the  same 
method  is  used  in  finding  the  H.  0.  F.  of  two  algebraic 
expressions. 

Find  the  greatest  common  factor  of  18  and  48. 

18)48(2 
36 

12)18(1 
12 
6)12(2 
12 

Since  6  is  a  factor  of  itself  and  of  12,  it  is,  by  (2),  a  fac- 
tor of  6  +  12,  or  18. 

Since  6  is  a  factor  of  18,  it  is,  by  (1),  a  factor  of  2  X  18, 
or  36 ;  and,  therefore,  by  (2),  it  is  a  factor  of  36  +  12,  or  48. 

Hence,  6  is  a  common  factor  of  18  and  48. 

Again,  every  common  factor  of  18  and  48  is,  by  (1),  a 
factor  of  2  X  18,  or  36 ;  and,  by  (2),  a  factor  of  48  -  36, 
or  12. 

Every  such  factor,  being  now  a  common  factor  of  18  and 
12,  is,  by  (2),  a  factor  of  18  --  12,  or  6. 

Therefore,  the  greatest  common  factor  of  18  and  48  is 
contained  in  6,  and  cannot  be  greater  than  6.  Hence  6, 
which  has  been  shown  to  be  a  common  factor  of  18  and  48, 
is  the  greatest  common  factor  of  18  and  48. 

124.  It  will  be  seen  that  every  remainder  in  the  course 
of  the  operation  contains  the  greatest  common  factor  sought ; 
and  that  this  is  the  greatest  factor  common  to  that  remain- 
der and  the  preceding  divisor.     Hence, 

The  greatest  common  factor  of  any  divisor  and  the  corre- 
sponding dividend  is  the  greatest  common  factor  sought. 


98  ALGEBRA. 

125.  Let  A  and  B  stand  for  two  algebraic  expressions, 
arranged  according  to  the  descending  powers  of  a  common 
letter,  the  degree  of  £  being  not  higher  than  that  of  A. 

Let  A  be  divided  by  B,  and  let  Q  stand  for  the  quo- 
tient, and  B  for  the  remainder.     Then 

B)AiQ 

BQ 

R 

Whence,  R  =  A- BQ,  ^nd.  A  =  BQ-\- R. 

Any  common  factor  of  B  and  R  will,  by  (2),  be  a  factor 
oi  BQ-\-  R,  that  is,  of  A  ;  and  any  common  factor  of  A  and 
B  will,  by  (2),  be  a  factor  oi  A  —  BQ,  that  is,  of  R. 

Any  common  factor,  therefore,  of  A  and  B  is  likewise 
a  common  factor  of  B  and  R.  That  is,  the  common  fac- 
tors of  A  and  B  are  the  same  as  the  common  factors  of  B 
and  R ;  and  therefore  the  H.  0.  F.  of  B  and  R  is  the 
H.  0.  F.  of  A  and  B. 

If,  now,  we  take  the  next  step  in  the  process,  and  divide 
B  by  R,  and  denote  the  remainder  by  8,  then  the  H.  C.  F. 
of  8  and  R  can  in  a  similar  way  be  shown  to  be  the 
same  as  the  H.  0.  F.  of  B  and  R,  and  therefore  the  H.  C.  F. 
of  A  and  B ;  and  so  on  for  each  successive  step.     Hence, 

The  H.  C.  F.  of  any  divisor  and  the  corresponding  divi- 
dend is  the  H.  C.  F.  sought. 

If  at  any  step  there  is  no  remainder,  the  divisor  is  a  fac- 
tor of  the  corresponding  dividend,  and  is  therefore  the 
H.  C.  F.  of  itself  and  the  corresponding  dividend.  Hence, 
the  last  divisor  is  the  H.  0.  F.  sought. 

Note.  From  the  nature  of  division,  the  successive  remainders  are 
expressions  of  lower  and  lower  degrees.  Hence,  unless  at  some  step 
the  division  leaves  no  remainder,  we  shall  at  last  have  a  remainder 
that  does  not  contain  the  common  letter.  In  this  case  the  given 
expressions  have  no  common  factor. 


COMMON   FACTORS   AND   MULTIPLES.  99 

Find  the  H.  0.  F.  of  Ix"  +  a;  -  3  and  4a;'  +  8a;'  -  a; -  6. 

2a;^  +  ^-3)4a;'  +  8a;'-    a;-6(2>a;  +  3 
4a;' +  2a;' -6a; 

6a;' +  5a; -6 
6a;'  +  3a;-9 

2a;  +  3)2a;'+    a;-3(a;-l 
2a;'  +  3a; 

-2a;-3 

.-.  the  H.C.F.  =  2a;  +  3.  -2a; -3 

Each  division  is  continued  until  the  first  term  of  the  remainder  is 
of  lower  degree  than  that  of  the  divisor. 

126.  This  method  is  of  use  only  to  determine  the  com- 
pound factor  of  the  H.  0.  F.  Simple  factors  of  the  given 
expressions  must  first  be  separated  from  them,  and  the 
H.  C.  F.  of  these  must  be  reserved  to  be  multiplied  into  the 
compound  factor  obtained. 

Find  the  H.  0.  F.  of 

12  a;*  +  30a;' -72  a;'  and  32  a;' +  84  a;' -  176  a;. 

12a;*  +  30a;'  -  72a;'  =  6a;'(2a;'  +  5a;  -  12). 
32a;'  +  84a;'  -  176a;  =  4a; (8a;'  +  21  a;  -  44). 
6  a;'  and  4  a;  have  2  a;  common. 

2a;'  +  5a;  -  12) 8a;'  +  21  a;  -  44(4 
8a;' +  20a; -48 

x-\-    4)2a;'  +  5a;-12(2a;-3 
2a;'  +  8a; 

-3a;-12 
.-.  theH.C.F.  =  2a;(.r  +  4).  -3a; -12 


100  ALGEBKA. 

127.    Modifications  of  this  method  are  sometimes  needed. 

(1)  FindtheH.aF.of4r^-8.2;-5andl2^'^-4:r-65. 

4:x''-8x-6)l2x''-    4a; -65(3 
12a:'' -24a: -15 


20  a: -50 

The  first  division  ends  here,  for  20a;  is  of  lower  degree  than  ix^. 
But  if  20a;  — 50  is  made  the  divisor,  4:X^  will  not  contain  20a;  an 
integral  number  of  times. 

The  H.  C.  F.  sought  is  contained  m  the  remainder  20a;  — 50,  and  is 
a  compound  factor.  Hence  if  the  simple  factor  10  is  removed,  the 
H.C.  F,  must  still  be  contained  in  2a;  — 5,  and  therefore  the  process 
may  be  continued  with  2  a!  —  5  for  a  divisor. 

2a;-5)4a:2-    8a:-5(2a;  +  l 
4a:2-10a; 

2a:-5 
2a:-5 


.-.  theH.C.F.-2a:-5. 

(2)  Find  the  H.  C.  F.  of 
21a;^  -  4.2:^  -  15a:  -  2  and  21a:'  -  32a:^  -  54a:  -  7. 
21a:' -  4a:^  -  15a:  -  2)21a:' -  32a:^  -  54a;  -  7(1 
21a;'-    4:x'-15x-2 


28  a;' -39  a;  — 5 


The  difficulty  here  cannot  be  obviated  by  removing  a  simple  factor 
from  the  remainder,  for  —  28 a;'^  —  39a;  —  5  has  no  simple  factor.  In 
this  case,  the  expression  21  a;^  —  4a;''  —  15a;  —  2  must  be  multiplied  by 
the  simple  factor  4  to  make  its  first  term  exactly  divisible  by  —  28  x^. 

The  introduction  of  such  a  factor  can  in  iw  way  affect  the  H.  C.  F. 
sought,  for  4  is  not  a  factor  of  the  remainder. 

The  signs  of  all  the  terms  of  the  remainder  may  be  changed  ;  for 
if  an  expression  A  is  divisible  by  —  F,  it  is  divisible  by  +  F. 

The  process  then  is  continued  by  changing  the  signs  of  the  re- 
mainder and  multiplying  the  divisor  by  4. 


COMMON   FACTORS   AND   MULTIPLES.  lOl 

28a;2  +  39a;  +  5)84a;'-    16x'-    60a:-    8(3a; 
84a;3  +  117a;'+    15a: 


-133a.'^-    75a:-    8 

Multiply  by  —  4, 

-4 

5320:^^  + 300  a: +  32(19 

532a:'^  +  741a:  +  95 

Divide  by  -  63, 

-63) -441a: -63 

7a:  +    1 

7a:  +  1)  28a:^  +  39a:  +  5(4a:  +  5 
28a:' 4-    4a: 


35a:  4- 5 
theH.C.F.  =  7a:+l.    35a:  +  5 


(3)  Find  the  H.  0.  F.  of 

8a;' +  2a: -3  and  6a:' +  5a:' -2. 


6a:'+    5.T'-    2 
4 

?'  +  2a:-3)24a-^  +  20a:-^-    8   (3a:+7 
24a:' +    6a:'-    9a: 

Multiply  by  4, 

14a:' +    9a:-    8 
4 

56.r'  +  36a:-32 
56  a:' +  14a: -21 

Divide  by  11, 

ll)22a:-ll 

2a:-    l)8a:'+2a:-3(4a:+3 
8a;'-4a: 

.-.  the  H.  C.  F. : 

6a:-3 
=  2a;-l.                             6a:-3 

102 


ALGEBRA. 


The  following  arrangement  of  the  work  will  be  found 
most  convenient : 


8^:^  +  2^-3 

6:r'+    5x'-    2 

8a;^-4rr 

4 

Qx-S 

24a;^  +  20:i;^-    8 

3x       ' 

6x-S 

24a;' +    6x'--    9x 

-    8 

Ux'-\-    9x- 

4 

b(jx'-i-S6x- 

-32 

+  7 

5Qx'+Ux- 

-21     ' 

ll)22a;- 

-11 

2x- 

-    1 

4a; +  3 

128.  From  the  foregoing  examples  it  will  be  seen  that,  in 
the  algebraic  process  of  finding  the  H.  0.  F.,  the  follow- 
ing steps,  in  the  order  here  given,  must  be  carefully- 
observed  : 

I.  Simple  factors  of  the  given  expressions  are  to  be  re- 
moved from  them,  and  the  H.  0.  F.  of  these  is  to  be  reserved 
as  a  factor  of  the  H.  0.  F.  sought. 

II.  The  resulting  compound  expressions  are  to  be  ar- 
ranged according  to  the  descending  powers  of  a  common 
letter ;  and  that  expression  which  is  of  the  lower  degree  is 
to  be  taken  for  the  divisor ;  or,  if  both  are  of  the  same 
degree,  that  whose  first  term  has  the  smaller  coeflicient. 

III.  Each  division  is  to  be  continued  until  the  remainder 
is  of  lower  degree  than  the  divisor. 

IV.  If  the  final  remainder  of  any  division  is  found  to 
contain  a  factor  that  is  not  a  common  factor  of  the  given 
expressions,  this  factor  is  to  be  removed;  and  the  resulting 
expression  is  to  be  used  as  the  next  divisor. 

V.  A  dividend  whose  first  term  is  not  exactly  divisible 
by  the  first  term  of  the  divisor,  is  to  be  multiplied  by  such 
a  number  as  will  make  it  thus  divisible. 


COMMON   FACTORS  AND   MULTIPLES.  103 

Exercise  42. 
Find  the  H.  0.  F.  of    - 

1.  5x^-\-4:x-l,  20a;' +  21a; -5. 

2.  2a;3~  4a;' -13a; -7,  6a:' -11  a;' -37a; -20. 

3.  6a*  +  25a'-21a2+4a,  24a*+ 112a'- 94a'  + 18a. 

4.  9a:'  +  9a;'-4a;  — 4,  45 a;' +  54 a;' -20 a;  — 24. 

5.  27a;«-3a;*  +  6a;'-3a;',  162a;«  +  48a;'- 18a;' +  6a;. 

6.  20a;' -  60a;' +  50a; -20,  32 a;* -92 a;' +  68 a;' -24 a;. 

7.  4a;' -8a; -5,  12a;'- 4a;  -  65. 

8.  3a'  — 5a'a;  — 2aa;',  9a'- 8 a'a;  -  20 aa;'. 

9.  10.^'  +  a;2-9a;  +  24,  20a;*- 17a;' +  48 a;  -  3. 

10.  8a;' -4a;' -32a; -182,  36a;'- 84a;' -  Ilia;- 126. 

11.  5a;'(12a;'  +  4a;'+17a;-3),  10a;(24a;'-52a;'+14a;-l) 

12.  9a;V-a;'3/'-20a;y*,  18a;'y- 18a;'2/' -  2a;2/'- 8y*. 

13.  6a;'  — a; -15,  9a;' -3a; -20. 

14.  12a;' -  9a;' +  5a; +  2,  24a;' +  10a;+ 1. 

15.  6a;' +  15a;' -6.^ +  9,  9a;' +  6a;' -  51a; +  36. 

16.  4a;'-a;'y-a;y'-52/',  7  x' i- 4:x'y  +  ixy"  -  Sy\ 

17.  2a3-2a'-3a-2,  3a'- a' -  2a- 16. 

18.  12y'+2y'-94y-60,  48y' -  24?/' -  348y  +  30. 

19.  9a;(2a;*-6a;'-a;'  +  15a;-10), 

6a;'(4a;*  +  6a;3  -  4a;' -  15a;  -  15). 

20.  15a;*+2a;'-75a;'+5a;+2,  3 5 a;* + or'- 1 75 a;'+ 30. 2;+ 1. 

21.  21  a;* -4a;' -15a;' -2a;,  21a;' -  32a;' -  54a;- 7. 

22.  9  a;*y- 22  a;y-3a;/+10y^,  9a;V-6a;y+ar'y'— 25  a;/. 


104  ALGEBRA. 

23.  Qx'^-Ax^—Ux^-Sx'-Sx-l,     • 

4:X*  +  2x'-lSx'-\-Sx-5. 

24.  x'-ax'-a''x''-a'x~2a\  3x' -1  ax' -i-Ba'x  ~2a\ 

129.  The  H.  C.  F.  of  three  expressions  may  be  obtained 
by  resolving  them  into  their  prime  factors;  or  by  finding 
the  H.  C.  F.  of  two  of  them,  and  then  of  that  and  the  third 
expression. 

For,  if  A,  B,  and  C  are  three  expressions, 

and  I)  the  highest  common  factor  of  A  and  B, 
and  U  the  highest  common  factor  of  D  and  C, 

Then  D  contains  every  factor  common  to  A  and  B, 

and  j&  contains  every  factor  common  to  D  and  0. 

.'.  -E  contains  every  factor  common  to  A,  B,  and  G. 

Exercise  43. 

Find  the  H.  0.  F.  of 

1.  2x''  +  x-l,  x''  +  bx-{-4:,  af'+l. 

2.  /-y'-y+l,  3?/'^-2y-l,  y^-y'  +  y-l. 

3.  x^-4^x'-\-9x-10,x^-\-2x'-3x-}-20,x^-j-bx''-9x'\-db. 

4.  x'~1x'+16x-12,  ^a^-Ux'  +  ie>x, 

bc(^-10x'  +  7x-14:. 

5.  3/^-5y^  +  lly-15,  2/'-y'  +  32/  +  5, 

2y'-7y'+16y-15. 

6.  2x'  +  Sx-6,  Zx^-x-2,  2o^^x-Z. 

7.  rp^-l,  x''-x^~x-2,  23?-c^-x~Z. 

8.  x'-Zx  —  2,  2a;'  +  3:c'-l,  a:'  +  l. 

9.  12  ix'  -  y%  10  ix'  -  f),  8  (A  +  ^/)- 

10.  x'-}-x7/,  c^y-\-y\  x'  +  xY-\-y\ 

11.  2{x^y  —  xy"^),  %{x^y  —  x-f),  A{x^y  —  xy^),  5(x^y  —  xy^). 


COMMON   FACTORS   AND   MULTIPLES.  105 

130.  Common  Multiples.  A  common  multiple  of  two  or 
more  expressions  is  an  expression  which  is  exactly  divisible 
by  each  of  the  expressions. 

The  lowest  common  multiple  of  two  or  more  expressions 
is  the  expression  of  lowest  degree  that  is  exactly  divisible 
by  each  of  the  given  expressions.  Thus,  24  {x^  —  y^)  is  the 
lowest  common  multiple  of  3  (a;  —  y)  and  8(a;  +  y). 

We  use  L.  CM.  to  stand  for  "  lowest  common  multiple." 

To  find  the  L.  0.  M.  of  two  or  more  algebraic  expressions: 

Case  I. 

131.  When  the  Factors  of  the  Expressions  can  be  found  by 
Inspection. 

(1)  Find  the  L.  C.  M.  of  ^^a'b'  and  60a^^V 

42a362=2x3x7xa-"'x62; 
eOa^i*  =  2  X  2  X  3  X  5  X  a^  X  6*. 
The  L.  C.  M.  must  evidently  contain  each  factor  the  greatest  num- 
ber of  times  that  it  occurs  in  either  expression. 

.-.  L.  C.  M.  =:  2  X  2  X  3  X  7  X  5  X  a'  X  Z'*, 
=  420a^6\ 

(2)  Find  the  L.  C.  M.  of 

4a;"' -h  4a; -48,  6a;'- 48a;  +  90,  4a;'- 10 a;  — 6. 

4x2+  4a;-48  =  4(a;2  +  a;-12)  =  2  x  2(a;-3)(.'r  +  4); 
6x2  -  48a;  +  90  =  6(a;2  -  8a;  +  15)  =  2  X  3(x  -  3)(a.  _  5) . 
4x2  -  lOx  -    6  =  2(2x2 -  5x -  3)  =  2(x  -  3)(2x  +  1). 

.-.  L.aM.  =  2x2x3x(a;-3)(a;  +  4)(a;-5)(2a;-f-l). 

Hence,  to  find  the  L.  C.  M.  of  two  or  more  expressions : 

Resolve  each  expressio7i  into  its  simplest  factors. 

Find  the  product  of  all  the  diff&u'ent  factors,  taking  each 
factor  the  greatest  number  of  times  it  occurs  in  any  of  the 
given  expressions. 


106  ALGEBEA. 

Exercise  44. 
Find  the  L.  C.  M.  of 

1.  4:a'x,  6aV,  2ax\  6.    2a; -1,  4a;' -1. 

2.  18ax\  72ay^  12a;y.         7.    a  +  b,  a' -{- b\ 

3.  x\  ax-\-x\  8.    a;'  — 1,  a;' +  1,  x'  —  l. 

4.  a;' -  1,  a;' —  a;.  9.    a;' -  a;,  a;3_  ^    .^s  ^  ^ 

5.  a^~h\  a'i-ah.  10.    a;' —  1,  a;' -a:,  r' -  1. 

11.  2a +  1,  4a' -1,  8a'  +  l.  . 

12.  (a  +  ^')'.  a^-h\ 

13.  4(1  + a:),  4(1-0;),  2(1 -a;'). 

14.  x-l,  x'  +  x-\-l,  x^-l. 

15.  x'-y\  {x  +  y)\  {x-y)\ 

16.  a;'-y',  2>{x-y)\  l2{x'-{-f). 

17 .  6  (a;'  -f  a;2/),  8  {xy  -  f),  10  {x'  -  f). 

18.  a;2+5a;+6,  a;'  +  6a:  +  8. 

19.  a'  — a  — 20,  a2  +  a-12. 

20.  a;' +  11a; +  30,  :x^ -\-l2x'\-2>b. 

21.  a;'-9a;  — 22,  a;'— 13a:  +  22. 

22.  4a^(a'-3a^>  +  25'),  5a'(a'  +  «&  -  6^>'). 

23.  20(a;'-l),  24  (a;' -a;- 2),  16  (a;' +  a;  -  2). 

24.  12a;y(a;'-y'),  2a;'(a;+y)',  ^y\x-y)\ 

25.  (a  —  i)(5  —  c),  {b  —  c)(c  -  a),  (c  —  a)(a  —  ^). 

26.  (a  —  b)(a-c),  (b-a){b-c),  (c-d)(c-b). 

27.  a;'-4a;'  +  3a;,  a;*  +  a;' -  12a;',  a,-^  +  3 a;*  -  4 a;'. 

28.  x'y  —  xy\  Sx{x  —  yy,  Ay(x-y)\ 

29.  (a-i-by-(c+d)\  (a+cy-{b  +  d)\  {a-\- d)' - (b -]- cf . 

30.  (2a;-4)(3a;-6),  (a;-3)(4a;-8),  (2a;-6)(5a;- 10). 


COMMON    FACTORS   AND   MULTIPLES. 


107 


Case  II. 


132.  When  the  Factors  of  the  Expressions  cannot  be  found  by 
Inspection. 

In  this  case  the  factors  of  the  given  expressions  may  be 
found  by  finding  their  H.  0.  F.  and  dividing  each  expres- 
sion by  this  H.  0.  F. 

Find  the  L.  C.  M.  of 

6a;'  -  ll:rV  +  2?/'  and  9a;'  -  l^xf  -  %y\ 

^^-Wxhj  +22/3 

6rc3-    8x^-4  ay 


3a;22/  + 4x2/2 +  27/3 
3a;2y  + 4x2/2 +  22/3 


9x3 
2 

-22x2/2 

-    82/3 

18x3 
18x3 

-44x2/2 
-33x2.v 

-162/3 
+    62/3 

112/L 

33x2y- 

44xv2- 

22  y3 

3x2    _ 

4x2/  — 

22/2 

2x-2/ 


.-.  the  H.C.F.  =  3x2 -4x2/ -22/2. 
Hence,        6 x^  -  11  x22/  +  22/'  =  (2 x  - 2/)(3 x^-^xy-  2y\ 
and  9x3  -  22x2/2  -  82/'  =  (3x  +  42/)(3x2  -  4xy  -  22/2). 


•.  the  L.  0.  M.  =  {2x  -  y){Zx  +  4y)(3a;'^  -  4a;?/  -  2y^). 


133.    The  product  of  the  H.  C.  F.  and  the  L.  C.  M.  of  two 
expressions  is  equal  to  the  product  of  the  given  expressions. 

For,  let  A  and  B  denote  the  two  expressions,  and  D  their  H.  C.  F. 
Suppose  A  =  al),  and  B  =  hD. 

Since  D  consists  of  all  the  factors  common  to  A  and  B,  a  and  b 
have  no  common  factor,  and  L.  C.  M.  of  a  and  6  is  ah. 
Hence,  the  L.  C.  M.  of  aD  and  hD  is  ahD. 
Now,  A  =  aD,  and  B  =  hD. 

.:  AB  =  ahD\ 

.'.  ^-~  =  ahD  =  the  lowest  common  multiple. 
D 

Hence,  the  L.  C.  M.  of  two  expressions  can  he  found  by 
dividing  their  product  hy  their  H.  C.  F. 


108  ALGEBRA. 

134.  To  find  the  L.  0.  M.  of  three  expressions  A,  B,  C. 
Find  if,  tlie  L.  C.  M.  of  ^  and  ^ ;  then  the  L.  C.  M.  of  M, 
and  C  is  the  L.  0.  M.  required. 

Exercise  45. 

Find  the  L.  C.  M.  of 

1.  e>x'-x-2,  2lx^-l1x-\-2,  Ux^-^bx-l. 

2.  x""—!,  x'-\-2x-Z,  Qx'-x-2. 

3.  x^~21,  x''-lbx-\-?>^,  a^-2>x^-2x-^^. 

4.  5a;^  +  19a;-4,  10a;'  +  13a;-3. 

5.  12x''~{-xy  —  &y\  18a;'  + 18rt:y- 20y^ 

6.  x'  —  2x^-^x,  2a;*-2a;'  — 2a;  — 2. 

7.  12a;^  +  2:r-4,  12a;^-42:i;-24,  12  a;'- 28  a?- 24. 

8.  a;' -  6a;' +  11a; -6,  a;' -  9 a;' +  26 a;  -  24, 

a;' -  8a;' +  19a; -12. 

9.  a;'-4a',  a;3+2aa;'  +  4a'a;+8a',  a;'-2aa;'+4a'a;-8al 

10.  x^-\-2x'y-xy''-2i/,  x^ -2x'y -  xf-\-2y\ 

11.  l+p+i9',  l-p-\-p\  l+/+i?^ 

12.  (1-a),  (l-a)',  (I -of. 

13.  {a-\-cf-h\  {a-^hy-c\  (h  +  cy-a\ 

14.  3c'-3cV  +  ^2/^-y',  4c'-c'2/-3c3/'. 

15.  m'-8m4-3,  m«  +  3m^  +  w  +  3. 

16.  20w*  +  w'-l,  25w*  +  5n'-w-l. 

17.  5* -25^ +  ^'-85 +  8,  46' -125'+ 96-1. 

18.  2r^-8r*+12r'-8r'  +  2r,  8r^-6r»  +  3r. 


CHAPTER  IX. 
FRACTIONS. 

135.  An  algebraic  fraction  is  the  indicated  quotient  of 
two  expressions,  written  in  the  form  y. 

0 

The  dividend  a  is  called  the  numerator,  and  the  divisor  h 
is  called  the  denominator. 

The  numerator  and  denominator  are  called  the  terms  of 
the  fraction. 

136.  The  introduction  of  the  same  factor  into  the  divi- 
dend and  divisor  does  not  alter  the  value  of  the  quotient, 
and  the  rejection  of  the  same  factor  from  the  dividend  and 
divisor  does  not  alter  the  value  of  the  quotient. 

Thus  i2^3,|>ii?=.3,lA±|  =  3. 

4  2x4  4-^2 

Hence,  it  follows,  that 

The  value  of  a  fraction  is  not  altered  if  the  numerator  and 
denominator  are  both  multiplied,  or  both  divided,  by  the 
same  factor. 

Reduction  of  Fractions. 

137.  To  reduce  a  fraction  is  to  change  its  form  without 
altering  its  value. 

Case  I. 

138.  To  reduce  a  Fraction  to  its  Lowest  Terms. 

A  fraction  is  in  its  lowest  terms  when  the  numerator  and 
denominator  have  no  common  factor.  We  have,  therefore, 
the  following  rule : 


110  ALGEBRA. 

Mesolve  the  numerator  and  denominator  into  their  prime 
factors,  and  cancel  all  the  common  factors ;  or,  divide  the 
numerator  and  denominator  hy  their  highest  common  factor. 

Keduce  the  following  fractions  to  their  lowest  terms : 


(1) 

(2) 
(3) 
(4) 


bla'bc^       3xl9a^5c^       3  a 

a^  —  x^_{a  —  x)(a^  +  ax-\-  x^)  __  a^ -\- ax -j- a^ 
o?  —  x^  (a  —  x){a  -\-x)  a-{-x 

ft^  +  7  g  +  10  _  (^  +  5)(a  +  2)  _  a  +  5 
a'  +  ba  +  e       (a  +  3)(a  +  2)      a  +  S 

6x'-5x-e  _(2:?;-3X3a;  +  2)_3a;  +  2 
8x''~2x-lb      (2x-dX4:X  +  b)      4a; +  6* 


^     x'-2x'-j-4:X-S 

In  example  (5)  we  find  by  the  method  of  division  the 
H.  0.  F.  of  the  numerator  and  denominator  to  be  a;  —  1. 
The  numerator  divided  by  a;  —  1  gives  a^  —  Sx-\-l. 
The  denominator  divided  by  a;  —  1  gives  x"^  —  x-\-d. 

X'  -  4:X''  +  4:X  —  1  _x''  -  Sx  +  1 

"  x^-2x^  +  4:X-3       x'-x  +  S' 


Exercise  46. 
Reduce  to  lowest  terms : 

x^-1  „      x^  —  2x-S 


1. 


4:x{x+l)  0^^-100;  + 21 

a;'-9a;  +  20  ^  '  ^^-^_±l 

:i;'-7a;+12'  *    x'-}-x+l 


FRACTIONS.  Ill 


a^-f       '  '      a^-2x''-x  +  2  ' 

f  g^  +  l  19     6a:^-23a:^4-16^-3 


aJ'  +  2a'  +  2a 

+  1 

a^-a-20 

a'  +  a-l2 

x^-4x'  +  9x 

-10 

x^  +  2x'  —  Sx 

+  20 

0:^-5:^2+11 

x-lb 

20. 


8.    Z "-"    '        ^.  21. 


22. 


x*-2a^-x'-2x+l 
a'  -  a'b  -  a'b'  +  ab^' 


(a  +  by 


x'-x'-}-3x  +  5  a^-ab-2b'' 

10^    x'^x'y-^xj^-t  23.    SaKo^'-^^) 
x'^  —  x^y  —  xi/ —  y*  '   ^{a^b  —  ab'^f 

^^     a^  +  4a''-5  g^^   a' +  2ab  +  b'' -  c" 

*  a^  —  3  a  +  2  a^-\-ab  —  ac 

^^      ^x'-\-2x-\  25.    6:r^-lla;^y  +  3a.y 
o;^  +  a;"^  —  a;  —  1     .  ^x^y  —  bxy"^  —  Qy^ 

a?-2,x'Jr^x-2  a^-(6  +  g  +  c^y 

•  a;'-a;^-2.^  +  2  *    {a-bf-^c  +  dy 

14     4ar'— 12aa:  +  9ft^  2^      6a;'-5rP-6 


8a;^-27a^  8a;2~2a;-15 

15a'^  +  a5-25^  rr^  +  a;y  +  y* 

9a*  +  3a6-2^>^  *    {x-y){x^-f) 

16^    a'-5^-2^>g-^^  29.    _^'+J/' 


a^  +  2a6  ■\-b''-c'  X*-  xy  +  y* 

x'-x'-2xi-2  Q     (a»  +  5^)(a'  +  a^>  +  y) 

2ar»-a;-l     '  '    (a'-J>'Xa'-ab-j-b') 


112  ALGEBRA. 


Case  II. 


139.   To  reduce  a  rraction  to  an  Integral  or  Mixed  Expression. 

x^  —  1 
(1)  Reduce to  an  integral  expression. 

X         i. 

^^  =  a;^  +  rr  +  l.     (§88) 


-1 


(2)  Reduce —  to  a  mixed  expression. 

X  -{-1 


x'-l[ 
x^  +  x" 

x  +  1 
x'-x  +  l 

- 

-0?- 

-1 

—  X 

x-1 

x  +  1 

tl 

-1_ 

1    1 

=  x'- 

-2 

2 

1     n 

X  +  1  x-\-l 

Note.    By  the  Law  of  Signs  for  division, 

■^and         2       =-- A_. 

03+1  —{x+1)  X  +  1 

The  last  form  is  the  form  usually  written, 

140.  If  the  degree  of  the  numerator  of  a  fraction  equals 
or  exceeds  that  of  the  denominator,  the  fraction  may  be 
changed  to  a  mixed  or  integral  expression  by  the  following 
rule: 

Divide  the  numerator  hy  the  denominator. 

Note.  If  there  is  a  remainder,  this  remainder  must  be  written  as 
the  numerator  of  a  fraction  of  which  the  divisor  is  the  denominator, 
and  this  fraction  with  its  proper  sign  must  be  annexed  to  the  integral 
part  of  the  quotient*!^ 


FRACTIONS.  113 


Exercise  47. 

Change  to  integral  or  mixed  expressions : 

^     x'-2x-{-l  g  I0a'-17ax-hl0x' 

X  —  1  ^a  —  x 

2.    3a;'^  +  2a;+l  ^  16 (3 .^^^  +  1) 

a;  +  4  *        407—1 

^  +  4  '         x  —  \' 

4     <^'  —  Q^^  +  a;'  q     a' +  5' 
a  +  a;  a  — J 

g     2a;^  +  5  ^^^      5^3  ,^^5 
a;-3  *  *   6a;'  +  4a:— 1' 


Case  III. 

141.   To  reduce  a  Mixed  Expression  to  a  Fraction. 

-     The   process  is  precisely  the   same   as  in  Arithmetic. 
Hence, 

Multiply  the  integral  expression  hy  the  denominator,  to 
the  product  add  the  numerator,  and  under  the  result  write 
the  denominator. 

Reduce  to  a  fraction  a  —  h  —  ^  ~"  ^   ~ — 

a-\-b 

a  +  b  a  +  b 

_a^-b'^-a^  +  ab  +  b^ 

a  +  b 
_^    ab 
a  +  b 

Note.  The  dividing  line  between  the  terms  of  a  fraction  has  the 
force  of  a  vinculum  affecting  the  numerator.  If,  therefore,  a  minus 
sign  precedes  the  dividing  line,  as  in  Example  (2),  and  this  line  is 


114  ALGEBRA. 

removed,  the  numerator  of  the  given  fraction  must  be  inclosed  in  a 
parenthesis  preceded  by  the  minus  sign,  or  the  sign  of  every  term  of 
the  numerator  must  be  changed. 


Exercise  48. 
Change  to  fractional  form : 

1.  i._^Zl^.  11.   ^^_(^_|.3^). 

x-\-y  07  +  3/ 

2.  I+^LZ^.  12.    ^^~^^^+6a  +  3rr. 

x-^y  4 

3.  3a;_l±M.  13.    a-l+-J_. 

X  a-\-\ 

4.  a  —  x-\ ■ 14.    x-\-b 

a  —  X  a;  —  3 

5.  ba—Ao — •     15.    2a  — 6  — 


5a  —  66  a-\-h 

6.    „  +  J_^±|!.  16.    3^-10+    " 


a  +  6  .         a;-f  4 


2 


7.  7^_2-3a  +  4a^  ^^^    ^^  +  ^+1  , 

5  —  6a  x—\ 

8.  3^_5aa:-3  ^g^    ^v_  3^ ._  3^13-^^ 

2a  a;-2 

9.  ^+4+1-  19-    a'-2aa;  +  4a;'^ ^^. 

a  — 6  a4-2a: 

10.  ^-1.  20.  x-g+y+^'-^y+y'. 

a  +  0  a:  +  a 


FRACTIONS.  115 

Case  IV. 

142.  To  reduce  Fractions  to  their  Lowest  Common  Denominator. 

Since  the  value  of  a  fraction  is  not  altered  by  multiply- 
ing its  numerator  and  denominator  by  the  same  factor 
(§  136),  any  number  of  fractions  can  be  reduced  to  equiva- 
lent fractions  having  the  same  denominator. 

The  process  is  the  same  as  in  Arithmetic.  Hence  we 
have  the  following  rule : 

Find  the  loiuest  common  multiple  of  the  denominators; 
this  will  he  the  required  denominator.  Divide  this  denomi- 
nator hy  the  denominator  of  each  fraction. 

Multiply  the  first  numerator  hy  the  first  quotient,  the  sec- 
ond numerator  hy  the  second  quotient,  and  so  on. 

The  products  will  he  the  respective  numerators  of  the 
equivalent  fractions. 

Note.  Every  fraction  should  be  in  its  lowest  terms  before  the 
common  denominator  is  found. 

Eeduce  ,  -— — -•  to  equivalent  fractions 

having  the  lowest  common  denominator. 

1  1 

x^  +  bx  +  Q    x'^  +  2x  +  l 

1  1 


(x  +  3)(x  +  2)     (a;  +  l)(x  +  l) 
.*.  the  lowest  common  denominator  (L.  C.  D.)  is 

(a;  +  3)(x  +  2)(ar  +  l)». 
The  respective  quotients  are 

(x  +  l)2and(a;  +  3)(x  +  2). 
The  respective  products  are 

(a;  +  l)2and(a;  +  3)(a;  +  2). 
Hence  the  required  fractions  are 

{x  + 1)^  ^^^         {x  +  mx  +  2) 

{x  +  3)(a;  +  2){x  +  1)'^  (x  +  2,){x  +  2)(x  +  1)' 


116  ALGEBRA. 


Exercise  49. 


Express  with  lowest  common  denominator : 

,     Sx-7    ^x-9  g  1  1 


2. 


6     ' 

18 

2x  —  4:y 

3a;- 

-8y 

bx'     ' 

10 

x 

4a  — 5c 

3a- 

-2c 

bac 

12a'c 

5 

6 

{a-h){h-cy  {a-b){a-c) 
«         4  a.-'  xy 


3.    ^^..       \     7,        •     7. 


8a;+2    2a:-l     3a;  +  2 
x-2'  Sx-6'  bx~l6 


Q     a  —  bTU    -I    c  —  bn 
a;    1~  X*  TUX  nx 


Addition  and  Subtraction  of  Fractions. 

143.  The  algebraic  sum  of  two  or  more  fractions  which 
have  the  same  denominator  is  a  fraction  whose  numerator 
is  the  algebraic  sum  of  the  numerators  of  the  given  frac- 
tions, and  whose  denominator  is  the  common  denominator 
of  the  given  fractions.  This  follows  from  the  distributive 
law  of  division. 

If  the  fractions  to  be  added  have  not  the  same  denomi- 
nator, they  must  first  be  reduced  to  equivalent  fractions 
having  the  same  denominator.     (§  142.) 

Hence,  to  add  fractions,  we  have  the  following  rule : 

Reduce  the  fractions  to  equivalent  fractions  having  the 
same  denominator ;  and  write  the  sum  of  the  numerators 
of  these  fractions  over  the  common  denominator. 

Note.  Each  fraction  should  be  expressed  in  its  lowest  terms. 


FRACTIONS.  117 

144.  When  tte  Denominators  are  Simple  Expressions. 

Simplify        3a-46_  20^-6  +  0^  a-4£. 
4  o  xA 

The  L.  C.  D.  =  12. 

The  multipliers,  that  is,  the  quotients  obtained  by  dividing  12  by 
4,  3,  and  12,  are  3,  4,  and  1. 

The  products  are 

9a-126,  8a-4&  +  4c,  and  a-4c. 

Hence  the  sum  of  the  fractions  equals 

9a-12b     8a-46  +  4c  .  a-4c 
12 


12 

'      12 

9a 

-126- 

-(8a 

—  46  +  4  c)  +  a 

-4c 

12 

9a 

-126- 

-8a  +  46-4c  +  a- 

Ac 

12 

2a 

-86- 
12 

Sc 

a  — 

46-4 

c^ 

6 

The  above  work  may  be  arranged  as  follows : 

The  L.  C.  D.  =  12. 

The  multipliers  are  3,  4,  and  1,  respectively. 

3 (3 a  — 46)     =     9a  — 126         =  1st  numerator. 

—  4(2 a  —  6  + c)  =  —  8a+   46  —  4c  =  2d  numerator. 

l(a  — 4c)        =        a  — 4c  =  3d  numerator. 

2a-   86-8c 
or      2  (a  —  4  6  —  4  c)  =  the  sum  of  the  numerators. 

e  c     L-  2(a  — 46  — 4  c)     a  — 46  — 4c 

.'.  sum  of  fractions  =  -^^ '  => 

12  6 


118  ALGEBRA. 

Exercise  50. 


Simplify : 


1     3a; -2y  .  bx-ly     8>r  +  2y 
bx      "^      \0x     "^       25 


2     4a:^-7.y'     3a;-8y     5-2.y 
3a;^  6a;  12 


4a'^+56''     3^  +  25  ,  l-2a 
26^^      "^      55      "^      9     * 


4.    ' —  ' 


bx         Vlx"^ 


7        "^      14  21       "*"      42      * 

g     3a;y-4      5y'+7      6a:'^-ll 
a;y  a:y^  rr^y 

^     a^-2ag  +  g'     ^>^-25g  +  c^ 
Q     5a'  — 2     3a' -a 

o. • 


Q     a  —  5  ,  5  —  c  ,  c  —  a  ,  aJ'  +  5c^  +  ca' 
cab  abc 


10. 


^^ 1 1     .  2a;-z  ■  y~-2'. 

2xhj     e>yh     2xz'       4a;V        40:^2 


FRACTIONS.  119 

145.   When  the  Denominators  have  Compound  Expressions. 

(1)  Simplify  ^-^—^^-^,-^,-. 

TheL.C.D.  is(a-6)(a  +  &). 

The  multipliers  are  a  +  6,  a  —  b,  and  1,  respectively. 

(a  +  6)(2  a  +  6)  =     2  a''  +  3  a6  +  6^  =  1st  numerator. 

-  (a  -  6)(2a  -  b)  =  - 20^  +  Sab  -  b^  =  2d  numerator. 

—  I(6a6)  =  —Qab  =  3d  numerator. 

0  =  sum  of  numerators. 

.'.  sum  of  fractions  =  0. 

(2)  Simplify  |5|  +  |5|  +  |5|. 

The  L.  C.  D.  is  {x  -  2){x  -  3){x  -  4). 
(x  -  l){x - 3)(x  -4)=    ar*-   8a;2  +  19a;-12  =  lst  numerator. 
[x  -  2){x  -  2){x  -  4)  =    a^-   8  a;^  +  20  »  -  16  =  2d  numerator, 
(x  -  2)(a;  -  S)(x  -  3)  =    a^  -   8  a;^  +  21  a;  - 18  =  3d  numerator. 

3  a^  -  24  a;2  +  60  ic  —  46  =  sum  of  numerators. 

»-     ,.  3x3 -24x2  + 60a; -46 

.•.  sum  of  fractions  = 


(x  -  2)(x  -  3)(a;  -  4) 
Exercise  51. 


Simplify : 
1.   A;  +  ^-  6. 


x-~6     x-\-b  2a(a-\-x)     2a{a  —  x) 

11  „  a  b 


x—1     x  —  S  {a-\-b)b     {a  —  b)a 

3.    -J-  +  ^.  8.  5  3 


1+a;      1-a;  2a;(a;  — 1)  ^x{x—2) 

4  _l 2_  g         1  +  ^  l-x 

1-x      l-a^  '    l-\-x^x'  \  —  x-\-3^ 

5  1       ■        X  jQ     2ax-Uy  2ax-\-Zby 
'   x-y     {x  —  yf  '   2xy{x-y)  2xy{x  +  y) 


120  ALGEBRA. 

Exercise  52. 
Simplify  : 

T         l,l|2a         _        X           x^      .       X 
1.    _— ^_ j- .      ^. ^ 


1-fa  1  —  a  1-a^             1-a;     1  —  a;      1  +  a;^ 

2.        ^              1       J  2:r  ^^    ^  J      y       J        a;' 

1— a;  1  +  a;  l  +  o;^            y     ^r  +  y     x^ -\- xy 

x  —  2  a;  —  3  a;  — 4 

g     _3 ,       4a  5a' 


7. 


8. 


x  —  a     (x  —  ay     (x~  ay 
1  1  3 


x~l     x-i'2     (x  +  lXx+2) 

a  —  b  ,  b  ~  c  ,  c —  a 


(b  +  cXc  +  a)      (c  4-  aXa  +  5)  '  (a  +  bXb  +  c) 


_     g;  —  CT  .  a;  —  5  (a  — by 

X  —  b     x  —  a     (x  —  d){x  —  b) 

10.    a;  +  y        2a;     ^     a;^  -  a;^ 

y       ^  +  y    y(^'  — y') 

11  <^  +  ^  I ^  +  g  ,  c-\-a 

(b  —  c)(c  —  a)      (c  —  a)(a  —  b)      (a  —  b){b  —  c) 

12  cb^  —  be         ■         b^  —  ac        .         g^  +  ct5 

-„        a  X  <^-\-x^ 

a  —  x     a-\-2x     (a  —  a;)(a  +  2  a;) 

14.    § 4 + 6 

{a  —  5)(5  —  c)      (a  —  b){a  —  c)      {a  —  c)(b  —  c) 


FRACTIONS.  121 


15. 
16. 


x~2y  _  2x-\-y  _     2a? 
x{x-y)     y{x-^y)     x'-f 

a  —  h a  —h    _  {a  —  b)(x  +  y) 

x(a  +  b)     y{a-{-b)         xy{a-\-b) 


17.        3rr  x-{-2y  ^       3y 

"  {x-\-  yf     x'  -y'      (x-  yf 


18. 


19. 


a  —  c  a 


-b 


{a-\-by-c'      (a  +  cy~b' 

a-{-  b  a  —  b     ,    ab(x  —  y) 

ax  -\-by     ax  —  by     aV  —  b^y^ 


146.  When  the  terms  of  the  denominators  are  not  arranged 
in  the  same  order. 

Since  —  =  a,  and  ~  ^  ■  —  a,  it  follows  that 
b  —b 

The  value  of  a  fraction  is  not  altered  if  the  signs  of  the 
numerator  and  denominator  are  both  changed. 

It  follows,  also,  by  the  Law  of  Signs,  that 

The  value  of  a  fraction  is  not  altered  if  the  signs  of  any 
even  number  of  factors  in  the  numerator  and  denominator 
of  a  fraction  are  changed. 

147.  Since  changing  the  sign  before  a  fraction  is  equiva- 
lent to  changing  the  sign  before  the  numerator  or  the 
denominator,  it  follows  that 

The  sign  before  the  denominator  may  be  changed,  provided 
the  sign  before  the  fraction  is  changed. 

Note.    If  the  denominator  is  a  compound  expression,  the  beginner 
must  remember  that  the  sign  of  the  denominator  is  changed  by 
changing  the  sign  of  every  term  of  the  denominator.     Thus, 
X      _  _      X 
a  —  X         X  —  a 


122  ALGEBRA. 

(1)  Simplify        l-^^  +  i^, 

Changing  the  signs  before  the  terms  of  the  denominator  of  the  third 
fraction,  and  the  sign  before  the  fraction,  we  have 
2 _      3      _  2x-3 
X     2x  —  \     4a;2  — 1 
The  L.  C.  D.  =  x{2x-  l){2x  +  1). 

2(2a;-l)(2a;  +  l)=     8x^-2    -  1st  numerator. 

—  3a;(2a;  +  l)         =  —  6a;'^  —  3a;  =  2d  numerator. 

—  a;(2x  — 3)  =- 2^^^  +  3 re  =  3d  numerator. 


sum  of  the  fractions  =  — 


2    =  sum  of  numerators. 

2 


x{2.x-l){2x  +  l) 
(2)  Simplify 

1,1,1 


a  {a  —  b)(a  —  c)      b(b  —  a){h  —  c)      c(c~  a)(c  —  b) 

Note.  Change  the  sign  of  the  factor  (6  —  a)  in  the  denominator 
of  the  second  fraction,  and  change  the  sign  before  the  fraction. 

Change  the  signs  of  the  two  factors  (c  —  a)  and  (c  —  b)  in  the  de- 
nominator of  the  third  fraction.     We  now  have 

.      I I + I 

a{a  —  b){a  -c)      b{a  —  b){b  —  c)      c{a  —  c){b  —  c) 
The  L.  C.  D.  =  abc  {a  -  b){a  -  c){b  -  c). 

be  {b  —  c)  =  b'^c  —  bc^  =  1st  numerator. 

—  ac{a  —  c)^  —  a^c  +  ac^  =  2d  numerator. 

ab  {a  —  b)=  a?b  —  aP  =  3d  numerator. 

a^b  —  a^c  —  ab''  +  ac^  +  b^c  —  bc^  =  sum  of  numerators. 

=  a2(5  _  g)  _  a(52  _  g2)  +  jc(5 _  c), 

=  [o?-a{b  +  c)  +  bc][b-c\ 

=  [a^  —  ab  —  ac  +  be]  [b  —  c], 

=  [{a'-ae)-{ab-be)][b-cl 

=  [a{a  —  c)  —  b{a  —  e)]  [b  —  c], 

^{a-b){a-c){b-c). 

.'.  sum  of  the  fractions  =     («-&)(«- c)(^> -A-  =  _L. 
abc  {a  —  b}{a  —  e){b  —  e)     abc 


FRACTIONS.  123 

Exercise  53. 


Simplify : 


X         X  —  y 


X  —  y  '  y  —  X 

2     3  +  2a;  .  3a;-2  .  \^x-x^ 
~2-x  "^  2  +  a:  "^    x' -  ^  ' 

„  X^         ,         X  X 


xi'  —  1      x-i-1      1  —  X 
4.        4_     ,^+      1 


5. 


S-Sy'      2-2y      6y  +  6 

1  w  2  ,  1 


(2  -  m)(3  -  m)      (m  -  l)(m  -  3)      (m  -  l)(m  -  2) 

1       +       ' 


{b-a)(x-{-a)      (a-bXx  +  b) 


a'  +  b'       2ab'    ,     2 


2a^5 


a^  '  a'^  +  ^' 

5  —  a _  g  —  25  _  3rg(a  —  &) 
'   x—b       b-i-x  h^  —  x^ 

g     3  +  2a:      2-3a;  .  l^x-x" 
2--X        2  +  x         x'-4:' 

3  7  4 -20a: 


10. 


l-2a;     l  +  2a;      4.r'-l 


11.  0^4-5  I  b-{-c  ■  c  +  a 


(6-c)(c-a)      (6-a)(a-c)      {a-b)(b-c) 

12  d^  —  be         .         5^  +  gg         .         c'  +  r?^ 

(a  ~  b){a -cy  Qi^  c){b  -ay  {c-  a){c  +  b) 


124  ALGEBRA. 


13.       y+^       [       ^  +  ^       I       ^  +  y 

{x-y){x-z)      {y-x){y~z)      {z-x){z-y) 


14. 


3  4  6 


(a  —  b){b  —  c)     {b  —  a){c  —  a)      {a  —  c){c  —  b) 


15,  1  ,  1 L. 

x{x  —  y){x  —  z)     y{y  —  x){y-z)     xyz 


Multiplication  and  Division  of  Fractions. 

148.   Multiplication  of  Practions. 

The  expression  f  X  ^  means  that  we  are  to  multiply  the 
quotient  -  by  a,  and  divide  the  result  by  b. 

From  the  nature  of  division,  if  we  multiply  the  divi- 
dend  c  by   a,   we   multiply   the   quotient  -   by  a,  and 

ac     .  .  .   . 

obtain  — ;  if  we  multiply  the  divisor  d  by  b,  we  divide 
ct 

the  quotient  ^  by  b,  and  obtain  ~     Hence, 
a  bd 

a     c ac_ 

b      d~bd 

Therefore,  to  find  the  product  of  two  fractions, 

Find  the  product  of  the  numerators  for  the  required 
numerator,  and  the  product  of  the  denominators  for  the 
required  denominator. 


In  like  manner, 

ace_ace_  ace 
b      d    f     bd     f     bdf 

and  so  on  for  any  number  of  fractions. 


FRACTIONS.  125 


^g--'  (7)=fxf  =  |- 


h    b    h 

In  like  manner, 

^Y=  — 
hi  ~~b^ 


149.  Division  of  Fractions.  If  the  product  of  two  numbers 
is  equal  to  1,  each  of  the  numbers  is  called  the  reciprocal  of 
the  other. 

The  reciprocal  of  7  is  -, 
0     a 

for  1x^  =  ^-^=1. 

a     0      ao 

The  reciprocal  of  a  fraction,  therefore,  is  the  fraction 
inverted. 

Since  f-?=l, 

0        0 

and  -  X  7  =  1,  it  follows  that 

a      0 

To  divide  hy  a  fraction  is  the  same  as  to  multiply  by  its 
reciprocal. 

To  divide  by  a  fraction,  therefore, 

Invert  the  divisor  and  multiply. 

Note.  Every  mixed  expression  should  first  be  reduced  to  a  frac- 
tion, and  every  integral  expression  should  be  written  as  a  fraction 
having  1  for  the  denominator.  If  a  factor  is  common  to  a  numera- 
tor and  a  denominator,  it  should  be  cancelled,  as  the  cancelling  of  a 
common  factor  hefore  the  multiplication  is  evidently  equivalent  to 
cancelling  it  after  the  multiplication. 


126  ALGEBRA. 


(1)  Fmd  the  product  of  —  X  ^  X  ^^, 

Za'b     6c''d      5ab^c  _2x  Qxba^h^d  _  ¥_ 
3c(P      5ab      Sah'^d^     3  X  5  X  8  a^Sc^c^*      2(P' 


(2)  Find  the  product  of 

^'  -  y'      V.  ^y  -^y""  X  ^'  -  ^y . 

x^  —  Sxi/-{-  2y^       x'^  +  ^y       (x  —  yy 


x^  —  3xy  +  2y^       x"^  +  xy       {x  —  yY 

{x-y){x  +  y)  ^^y{x-2y)^      x{x-y) 
{x-y){x-2y)       x{x  +  y)       {x-y){cc-y) 

y 


Note.     The  comiiion  factors  cancelled  are   x  —  y,  x  +  y,  x  —  2y, 
X,  and  x—y. 

(3)  Find  the  quotient  of 


(a  —  xf     d^  —  x^ 

ax      ^     ah      _  ax  (a  —  x)(a  +  z) 

(a  —  x)'^     a^  —  x^     (a  —  x){a  —  x)  ah 

_x{a  -\-  x) 
b{a  —  x) 

The  common  factors  cancelled  are  a  and  a  —  x. 

Exercise  54. 
Simplify : 

1     ^x—  3        ^^      •     ^^ 

'    bx      d  '    2p-2'  p-1 

2x     Sa^     Sac  ^      8x*y   ,    2r> 

a         c         2b'  '    Ibab'  '  Sab' 


FRACTIONS.  127 


5.    T^^r-X— -^-  10.— rX  — 


45  a;^      24  a'^'^  a^  +  ah     a^  -  ah 

^xYz          20a'b'c  ^j   a'  +  b',  a-b 

'    10 a'b^c          18xfz'  '  a'-b'  '  a  +  b 

.y     3^^5y^^_12^'  ^^x'-^x-2^x'~lSx+^2 

4iXz^      Qxy         2x1/*  x^  —  7x           x'^-\-2x 

g     9mV^5^^24^2  ^^   x^ -llx-^^O  ^,x^ -Zx 

8p^q^      2  xy     90  mn  x^  —  ^x-{-^       x^  —  bx 

25Fm''     lOn'g     Spm  ^^   a'  —  x^^Ja-h  xf 

'    14  nY     Ibp'm     ^khi  '  a^  ■\- x"     {a  -  xf 

15.  -Mpl^yh' X .      ^ 

ca;  (a;  -  y){x  +  y)' 

a^  +  2a5      ah -2b'' 
'    a^-{^U'      a^-W 


X 


17.     -^^ r^X 


4  .x'-25 


i8.-^'+^yx^ 

a;  —  y        a;*  - 

-  g   m^  —  r^  ,  ?i  - 

-  W, 

x^-^bx     x'i-2x  c^  +  d^       c-i-d 


a' -5a +  4      a'— 10a +  21      a2-5a 
^>^-7&  +  6     5^  +  105  +  24  .   b'  +  Qb 


22.  ^'  -  y'        X  ^y  -  2y'  V.  a;^  -  xy 

x"^  —  Sxy-{-2y'^       x^ -\- xy       (x  —  yy 

^„     a'-Ba'bj-Sab'-b'     2ab-2b'  ^  a'  +  ah 
23-  ^^3^^  =  3        ^1^^' 


128  ALGEBRA. 

(g  +  ^y-oV  c'-ja  +  hY 
'    {x-bf-a'     x'-(a-by 


(a  +  by-(c  +  d)\,  (a-cy-(d-by 

(a+<y-~(b  +  dy  '  (a-by--(d-cy 

x^  -^-^xy  -{-y"^  —  z^     x  —  y-\-z 


26. 


150.  Complex  Practions.  A  complex  fraction  is  one  that 
has  a  fraction  in  the  numerator,  or  in  the  denominator,  or 
in  both. 

Note.  Generally,  the  shortest  way  to  simplify  a  complex  fraction 
is  to  multiply  both  terms  of  the  fraction  by  the  L.  C.  D.  of  the  frac- 
tions contained  in  the  numerator  and  denominator. 

a-\-x     a  —  x 


(1)  Simplify  ^~^     ^  +  ^- 

Q,  ~\-  X    1^  CL         X 


a  —  X     a-\-x 

The  L.  C.  D.  of  the  fractions  in  the  numerator  and  denom- 
inator is 

(a  —  x){a  +  x). 

Multiply  by  (a  —  x){a  +  x),  and  the  result  is 

{a-\-xy-(a-xy 

{a  +  xy  +  {a-xy 
_{a^  +  2ax  +  x")  -(a'~2ax-{-  x') 

(a'  +  2ax  +  x')  +  (a'  -  2ax  +  x') 
^d^-\-2ax-\-x^-a^-\-  2ax  —  x"" 
~  a^-[-2ax  +  x'-\-o^-2ax-\-  o^ 
_      Aax 
~2a''  +  2x' 
__    2ax 
~d'-\-x''' 


FRACTIONS. 

(2)   Simplify 

X 

1 

X 

'+^+1-1+^ 

X 

X 

1                     ^ 

^              x{l-x  +  x') 

\      \     T     1 

X                    {l+x)(l~x  +  x')-{-x 

•»•   r  •*'  r  , 

-x-i-x" 

1      x-x'-j-x' 

x(l  +  x  +  x') 

l  +  x  +  a^-^x-x'  +  x') 

x-\-x'-{-x' 

l  +  x' 

129 


Note.     In  a  fraction  of  this  kind,  called  a  continued  fraction,  we 
begin  at  the  bottom,  and  reduce  step  by  step.     Thus,  in  the  last 

example,  we  take  out  the  fraction  ,  and  multiply 

1  —X  +  x^ 
the  numerator  and  denominator  by  1  —  x  +  a;'',  getting  for  the  result, 

^{^-^  +  ^') ,  which  simplified  is  ^-^'  +  < 

{I  +x){l-x  +  x'^)  +  x  1  +  x  +  x^ 

Putting  this  fraction  in  the  given  complex  fraction  for 


1+X  + 


l—x  +  x^ 


we  have 


X 


+  a^ 


1+x  +  a^ 
Multiplying  both  terms  by  1  +  a;  +  a;*,  we  get 
x(l  +x  +  ar^) 
1+x  +  x^  —  x  +  x^  —  a^ 

_x  +  x^  +  x^ 


130  ALGEBRA. 

Exercise  55. 
Simplify : 

Sx  ■  x—1 

1.  ^  ^,         .  8.    1 


X  —  a 


a;_(^-^)(^-^) 


1-: 

1 

x-y 

X  —  a 
X  -{-a 

X 

6 

x^-f 

X 

y 

xy  +  2/2 
^+1  , 

x^  +  ^y 

x-\ 

7 

x~\  ' 

x-\-\ 

rr+l 

x-\ 

13.„   ,    -,^_.'r       oi  1.1 

6 


'(^+l)-f-2i  1  + 

D  d  x 


X—  1  +  ~      /. 
2.    lii^.  9.    1  + 


a:-2  +  -^  \-\-x  '    ^^' 


a;  —  6  1  —  a; 


3.   -1 2^-^  ^, 


2     2  1  +  1 


11. 


1  + 


x  +  a  l  +  rr  +  ^il     ' 

1—  X 

\^^ aj\x      a)  \x      a       j\x      a       J 

\x     a) 


13. 


x~y 

(^^- 

-f){2:^- 

2xy) 

[4 

i{x-yf 

xy 

x  —  \      x-\-l  3c-\-y 


FRACTIONS.  131 

ab ac^ 

15     x^ -\-{^-\-b)^-\-CLb      x^  -\-{a-\-c)x-{'  ac 

b  —  c 

x'  +  {b-{-c)x-\-hc 

a-\-  b  ■       b 

16.    -i_4-l L_.  17.    _b__a±± 

1+1  ^+1  1+1 

X  '  a      b 

2m-3+i-  4  +  -  +  f 

18.    11.    19.    ^^      ^^      ^^     20.  ^ 

2m- 1  g^  -  (^  +  g)'  ^  •  3 


w  a6  -.   ,      3 

1  —  X 

Exercise  56. 

MISCELLANEOUS   EXAMPLES. 

1.  Simplify  ^--9.3  +  7.^  +  9.-8 

2.  Find-the  value  of  ""'^f  ""^'tof  ^^len  a  =  4,  6  =  J, 

■|.  a^--¥  —  c^-\-2bc 

3.  Find  the  value  of  3  a''  +  ^^-  -  -  when  a  =  4,  5  =  J, 

(7=1.  ^        ^' 

4.  Simplify  ? 1 L_. 

'6.    Find  the  valueof(^^-:=:^Y-^~^^  +  ^when.  =  ^ 
\x-bj      x  +  a  —  2b  2 

7.   Simplify  j-^-ti g-^    4,_^g_l« 


a  +  & 

-b 
2b 


132 


ALGEBRA. 


10. 

2 


Simplify  (^^±y^  -  ^-=^  ^  (^:±1  -  ^-^^ 
\x^-y'     x'Ary")      \x  —  y     x-\-y_ 

Simplify 

\y      JV~y 

Simplify 

-b')\      a-\-b      y\o?-\-h' 


/"^W'       J\x'-\-xy  +  f      V 


11 


a^  -{-ah-\-  h 


1     1^  Qt  —  X        -I     ,    (X         X 


11.    Simplify 


a  -j-  X 


a'  +  x' 


-.  _ a~ X      -J      d^  —  x'^ 


a-\-x 


a^-^x"- 


12. 


Divide  x'-\-\-%(--  x'\^dx-\-^\  by  a;  + 1- 
x^        \yr         J         \        xJ  X 


"2.  XI 


V  13.    Simplify  ^l±yl^. 


'^y 


{^-yj 


1  +  -^ 


14. 


15. 


16. 


Find  the  value  of  |±^  -f  ^     ^  "^ 


4aJ 


a5 


26-a;     2^^  +  a;     45'-a;^ 


when  a; 


the 
ah-\-a 


Find  the  value  of  ^ — -   when   a:=     "^       and 

a:  —  y+1  ao  +  1 

3/ 


ah^\ 


Simplify 
1 


a{a  —  h){a  —  c)      h{h~c){b  —  a)      c(c  —  a)(c  —  b) 


17.    Simplify 


Sabc 


a  —  1  ■  b  —  1  .  £_ 


6c  +  ca  —  a6 


a     0     c 


FRACTIONS.  133 


—  1^ 

2  /v,2 


18.    Simplify  — X 


11  m'  +  7i 


^   ^ 


19.   Simplify  -^ ^  1 1+       2&.       I  • 

a     5  +  c 

i20.   Simplify  3a-[6+f2a-(J-c)n+|  +  |^- 

1  1      I        a; y 

21.    Simplify  "-^     "-y     («-x)'      (g-y)' 


(a  -  y)(a  -  xj     (a  -  a;)(a  -  yf 


,2.   Simplify  L^.       23.   (^-.')(2^'-2^). 

3  —  2; 
24.   Simplify  (^_^-^^(^_  +  -,_^j. 

26.  Simplify  — r:r; ^+771 wl ^ T' 

^    "^   a(a  — 5)(a  — c)      b(b  —  a)(b  —  c)      ahc 

27.  Simplify  |±i  x  "^  ~  "^ 


^ 6_      '^{x-~l){x-2) 

x—\ 


CHAPTER  X. 

ise^tCnOTTAL   EQUATIONS.       ^^.^.^vV-V^f 
151.   To  reduce  Equations  containing  Tractions.         -^-^^  <^ 

(1)  Solve|-^  =  :i;-9. 

Multiply  by  33,  the  L.  C.  M.  of  the  denominators. 
Then,  na;-3x  +  3  =  33a;- 297, 

11  a;  -  3a;  -  33a;  =  -  297  -  3, 
-25a;  =  -300. 
.-.a; -12. 

Note.  Since  the  minus  sign  precedes  the  second  fraction,  in 
removing  the  denominator,  the  +  (understood)  before  x,  the  first 
term  of  the  numerator,  is  changed  to  — ,  and  the  —  before  1,  the 
second  term  of  the  numerator,  is  changed  to  +. 

Therefore,  to  clear  an  equation  of  fractions, 

Multiply  each  term  hy  the  L.  C.  M.  of  the  denominators. 

If  a  fraction  is  preceded  by  a  minus  sign,  the  sign  of  every 
term  of  the  numerator  must  be  changed  when  the  denoini- 
nator  is  removed. 

(2)  Solve  ^^-^^=^^-^^. 
^  ^  x-b     x-e>      x-S     x-9 

Note.  The  solution  of  this  and  similar  problems  will  be  much 
easier  by  combining  the  fractions  on  the  left  side  and  the  fractions 
on  the  right  side  than  by  the  rule  given  above. 

(a;  _  4)(a;  _  6)  -  (a;  -  5)^  _{x-  7){x  -  9)  -  (ar-  8)' 
(a;  _  5)(a;  -  6)  (a;  -  8)(a;  -  9) 


FRACTIONAL  EQUATIONS.  135 


By  simplifying  the  numerators,  we  have 
-1  -1 


(.c  _  5)(x  -  6)     {x-  8){x  -  9) 

Since  the  numerators  are  equal,  the  denominators  are  equal. 
Hence,  (x  -  h){x  -  6)  =  (x  -  8)(«  -  9). 

Solving,  we  have  x  =  7. 

Exercise  57. 

Solve :  ' 

3           3  DO 

,     5-2a;,o_  6a;-8     «     a:  +  2_14      3  +  5^; 

4  2                  2          9           4 

5a:+3  3  — 43;.a;_31      9  — 5a; 

8  3          2       2           6      ' 

8.  10£±-3-^£zi7^  10(^-1). 

9.  i£^-2^±l=.3:.-14. 

2  3 

7a;  +  5     5.r— 6_8-5a; 
6  4  12     ' 

11     ^  +  ^     a:  — 4_o  ,  3a:— 1 
^^-    -^^  5  ^  +  ~15~* 

12.  33;  +  5      23;+7  ^  ^q      ^^-Q. 
To  o 

13.  l(3a;-4)4-k5a;  +  3)=43-5ar. 
7  o 

14.  |(27-2:.)  =  |-^(7^-54). 


10. 


136  ALGEBRA. 


15.  bx—lSx  —  S[16-6x-(4:-bx)]l=6. 

16.  5^-3_9^-^^5^     19 

7  3  2       6^         ^ 


17. 
18. 
19. 


2^7+7  9a:-8_a;-ll 

7  11  2     ■ 
8a;— 15  11a;- 1  _  7^7  + 2 

3  7               13     ' 

7a; +  9  3a;  +  l_9a;-13     249 -9a; 

8  7  4                 14      ■ 


152.  If  the  denominators  contain  both  simple  and  com- 
pound expressions,  it  is  best  to  remove  the  simple  expres- 
sions first,  and  then  each  compound  expression  in  turn. 
After  each  multiplication  the  result  should  be  reduced  to 
the  simplest  form. 

Multiply  both  sides  by  14. 

Then,  8a;  +  5  + '^^'^^~  ^^  =  8a;  +  12. 

3a;  +  1 

Transpose  and  combine,     ^  '^~   -^  =  7. 
3a;-t- 1 

Divide  by  7  and  multiply  by  3  a;  +  1, 

7a;- 3  =  3a; +  1. 

.•.a;-L 

e)  ^X  I  X  ry 

(.)  Solve  ^  =  i- V- 

Multiply  both  terms  of  each  complex  fraction  by  9. 

Then,  27,^^1_7^^. 

36  4  90 

Solving  this  equation,  we  have  x  =  Q. 


FRACTIONAL    EQUATIONS.  137 

Exercise  58. 


Solve  the  equations : 

36  5^-4  "^4* 

9(2^--3)      lla;-l_9a;+ll 
14        ~^  Zx+1  7 


3. 


10a;+17       12a;  +  2_5a;-4 
18  13:r-16  9 

^     6:r+13       3:r  +  5  __2x 
15  5a; -25       5* 

^     18^-22^^^^1  +  16^^^         101-64, 


9. 


10. 


39-6a;   '        ^24  ^  24 

6-5a;        7-2:r=^    _l  +  3:r      10;r-ll 


15         14  (a; -1)         21  30        '  105 

^x  +  b  ,  8a;-7_36;r  +  15  ■  41 


8. 


14 

'  6a;  +  2            56 

Qx  +  1 

2a;-2__.2a:+  1 

15 

7a;— 6           5 

6:i;+l 

2a;-4       2a;-l 

15 

7a;- 16          5 

Ix-Q 

a;  — 5          X 

35  6a; -101      5 


163.  Literal  equations  are  equations  in  which  some  or 
all  of  the  given  numbers  are  represented  by  letters;  the 
first  letters  of  the  alphabet  are  used  to  represent  known 
numbers. 


138  ALGEBRA. 

(1)  {a-x){a-{-x)  =  2a^-\-2ax-x\ 

Then,  a? -x"  =  2a^  -^  2ax-  x\ 
—  2ax  =  a^. 
'  —  _  ^. 

(2)  (x  —  a)(x ~h)  —  {x  —  h){x  —  c)  =  2{x  —  a){a  —  c). 

(a;2  -ax-bx  +  ab)  —  {x^  —  hx  —  cx  +  bc)-=2  {ax  —  ex  -  a?  +  ac), 

x^-ax  —  bx  +  ab  —  x^  +  bx  +  cx  —  bc=  2ax  —  2cx  —  2a'^  +  2ac. 

That  is,  —  S ax  +  S ex  =  —  2 a^  +  2aG  —  ab  +  be, 

—  3{a  —  c)x  =  —  2a{a  —  c)  —  b{a  —  e), 

-3x  =  -2a-b. 

2a  +  b 
.■.x  =  ^-. 

Exercise  59. 
Solve  the  equations : 
1.    ax -}- be  —  bx -}- ac.  2.    2a  —  ex —  Sc  ~bbx. 

3.  a'^x  -\-  bx  —  c  =  b'^x  -\-  ex  —  d. 

4.  —  ac^  -\-  h^e  +  obex  =■  abe-\-  cinx  —  ae^x  +  b'^c  —  me. 

5.  {a-\-  x-{-  b){a  -\- b  —  x)  —  {a -\-  x){b  —  x)  —  ab. 

6.  (a'  +  xY'-^x'  +  ia'  +  a*. 

7.  (a"^  —  x){a^  -\-x)  —  a^-\-2ax  —  x^. 

^     ax  —  b  ,          X  -{-  ae         -^              ^a  —  bx      1 
8. |-a  =  — ■ 10.    ax — = -• 

e  c  *    2  z 

_     a(b'^x-\-x^)  ,  ax"^     --     n        ^ax—2b 

9.       ^    ^   ' ^-=aex-\---—    11.   6a —  x. 

bx  b  o 


12. 

x'  —  a      a  —  X _  Ax      a 
bx            b            b       X 

13. 

3      oh  —  x^  _^lX  —  ae 
e          bx               ex 

14. 

7        ax   ,    X        r\ 

am  —  0  — --  -j =0. 

0       m 

15 


16. 


FRACTIONAL    EQUATIONS.  139 

2»ax  —  2b      ax  —  a      ax      2 


u 

2b 

b 

3 

ab  -\-  X 

h'- 

-  X 

X  — 

b 

ab- 

X 

'        b' 

a 

'b 

a' 

b' 

bx^l_ 

a(x' 

-1) 

I 

9. 

ab 

he, 

17.  aa;-^^^^-I^  =  "'^"  ~  "A  19.    ^  =  66'H-c^  +  -- 

XX  X  X 

18.  _^?£:!_  +  a  +  25  =  0.  20.    "('^+^')  =  ac+gg. 
6  —  cx              (7  dx  d 


Exercise  60. 

I^olve  the  equations  : 

X  —  ?>    _    X  —  5     .  1 
4:{x-l)~Q{x-iyd 


1. 


2.    x\      ""     _{x-2){x  +  4:) 
x—1  X  -\-l 

3         7      _6a;+l      3(l  +  2■^•^) 
'    x—1       x-{-l  x'^  —\ 

1  1  X 


4. 


2(a;-3)      3(:r-2)      (a;-2)(a;-3) 


,       2(2:^  +  3)  _     6  5:r+l 

9(7 -a;)        1-x     4(7 -a:)' 

6.    _l^_4=^(21+M_io. 
a;  +  3  3a;  +  9 


rj     x—1      2a;— 15  1 


x+1       2x-^      2(a;+7) 

8      a:  +  4       p^3a;  +  8 
'   3a;  +  5'^    '      2a; +  3 


140  ALGEBRA. 


9     132^+1  ,  8a:  +  5_^o       ^^     Sx~l     4a:-2__l 
3a;  +  l         a;-l  '        ■  *    2^-1      3.^-2      6" 

10.    -A_+_L:  =  ^.      12        3        :.+  !__     ^^ 


2;r-3      x-2      3a;  +  2  :r-l     :i;-l      l~x' 

IS     ^~4      :r  —  5_a;  —  7      :g  —  8 
'   X  —  6      X  —  6      x-'8      X  —  9 

14.  (x  -  aXx  -b)  =  {x-a  —  h)\ 

15.  {a~b){x  —  c)-{b-c){x-a)-{c-a){x-b)  =  0. 

16.  ^lzi^±i  +  ^!+^+i  =  2:r. 

a;—  1  x-\-l 

17.  ^^       7  3^ 


a;  +  2      :r  +  3      x'  +  bx-{-& 
18.    (^^+l)2z=:a;[6-(l-a:)]-2. 

^g     2b-^x      Ux  +  ^_    23        ^ 

a;+l  3a;  +  2       a;+l        * 

„„      3a5c    ,       a^b"^       ,  (2  a +  ^)^'^:?;      o       .  hx 


21. 


29 


-8      2a: -16      24      3:r-24 


l2      :ry     2  4 


23. 


1 3     __    '^l^x 

5      :r-l  3 


24.       1 l_  +  _^ 3_  =  lj 


(-i) 


PROBLEMS.  141 

154.   Problems  involving  fractional  Equations. 

Ex.  The  sum  of  the  third  and  fourth  parts  of  a  certain 
number  exceeds  3  times  the  difference  of  the  fifth  and  sixth 
parts  by  29.     Find  the  number. 

Let  X  =  the  number. 

Then      -  +  -  =  the  sum  of  its  third  and  fourth  parts, 

-  —  -  =  the  difference  of  its  fifth  and  sixth  parts, 
5     6  ^ 

3|  E_  E  j  =3  times  the  difference  of  its  fifth  and  sixth  parts, 
-  +-  —  3(-  —  -1  =  the  given  excess. 

3    4      V5    ey         ^ 

But  29  =  the  given  excess, 

.-.  1^^^2>i--^  -29. 
3     4        V5     6/ 

Multiply  by  60,  the  L.  C.  D.    of  the  fractions. 

20x  +  15a;  -  36a;  +  30a;  =  60  X  29. 

Combining,  29  a;  =  60x29. 

.-.  a;  =  60. 

Exercise  61. 

1.  Find  the  number  whose  third  and  fourth  parts  together 

make  14. 

2.  Find  the  number  whose  third  part  exceeds  its  fourth 

part  by  14. 

3.  The  half,   fourth,  and  fifth  of  a  certain  number  are 

together  equal  to  76 ;  find  the  number. 

4.  Find  the  number  whose  double  exceeds  its  half  by  12. 

5.  Divide  60  into  two  such  parts  that  a  seventh  of  one 

part  may  be  equal  to  an  eighth  of  the  other. 

6.  Divide  50  into  two  such  parts  that  a  fourth  of  one 

part  increased  by  five-sixths  of  the  other  part  may 
be  equal  to  40. 


142  ALGEBRA. 

7.  Divide  100  into  two  such  parts  that  a  fourth  of  one 

part  diminished  by  a  third  of  the  other  part  may  be 
equal  to  11. 

8.  The  sum  of  the  fourth,  fifth,  and  sixth  parts  of  a  cer- 

tain number  exceeds  the  half  of  the  number  by  112. 
What  is  the  number  ? 

9.  The  sum  of  two  numbers  is  5760,  and  their  difference 

is  equal  to  one-third  of  the  greater.  What  are  the 
numbers  ? 

10.  Divide   45   into   two  such  parts  that  the  first  part 

divided  by  2  shall  be  equal  to  the  second  part  mul- 
tiplied by  2. 

11.  Find  a  number  such  that  the  sum  of  its  fifth  and  its 

seventh  parts  shall  exceed  the  difference  of  its  fourth 
and  its  seventh  parts  by  99. 

12.  In  a  mixture  of  wine  and  water,  the  wine  was  25  gal- 

lons more  than  half  of  the  mixture,  and  the  water 

5  gallons  less  than  one-third  of  the  mixture.  How 
many  gallons  were  there  of  each  ? 

13.  In  a  certain  weight  of  gunpowder  the  saltpetre  was 

6  pounds  more  than  half  of  the  weight,  the  sulphur 
5  pounds  less  than  the  third,  and  the  charcoal  3 
pounds  less  than  the  fourth  of  the  weight.  How 
many  pounds  were  there  of  each  ? 

14.  Divide  46  into  two  parts  such  that  if  one  part  be 

divided  by  7,  and  the  other  by  3,  the  sum  of  the 
quotients  shall  be  10. 

15.  A  house  and  garden  cost  $850,  and  five  times  the 

price  of  the  house  is  equal  to  twelve  times  the 
price  of  the  garden.     What  is  the  price  of  each? 


PROBLEMS.  143 

16.  A  man  leaves  the  half  of  his  property  to  his  wife,  a 

sixth  to  each  of  his  two  children,  a  twelfth  to  his 
brother,  and  the  remainder,  amounting  to  $600,  to 
his  sister.     What  was  the  amount  of  his  property  ? 

17.  The  sum  of  two  numbers  is  a  and  their  difference  is  b ; 

find  the  numbers. 

18.  Find  two  numbers  of  which  the  sum  is  70,  such  that 

the  first  divided  by  the  second  gives  2  as  a  quotient 
and  1  as  a  remainder. 

Hint.  Dividend -Remainder  ^  q^^^.^^^ 

Divisor 

19.  Find  two  numbers  of  which  the  difference  is  25,  such 

that  the  second  divided  by  the  first  gives  4  as  a 
quotient  and  4  as  a  remainder. 

20.  Divide  the  number  208  into  two  parts  such  that  the 

sum  of  the  fourth  of  the  greater  and  the  third  of  the 
smaller  is  less  by  4  than  four  times  the  difference  of 
the  two  parts. 

21.  Find  four  consecutive  numbers  whose  sum  is  82. 

Note.     If  x  represent  a  person's  age  at  the  present  time,  his  age 
a  years  ago  will  be  represented  by  a;  —  a,  and  a  years  hence  by  a;  +  a. 

Ex.    In  eight  years  a  boy  will  be  three  times  as  old  as  he 
was  eight  years  ago.     How  old  is  he  ? 

Let  X  =  the  number  of  years  of  his  age. 

Then  a;  —  8  =  the  number  of  years  of  his  age  eight  years  ago, 
and  X  +  8  =  the  number  of  years  of  his  age  eight  years  hence. 

Since  his  age  8  years  hence  will  be  three  times  his  age  8 
years  ago,  we  have 

x  +  8    =3(x-8), 

a; +  8    =3x^24, 

a;  -  3  a;  =  -  24  -  8, 

-2a;  =  -32, 

a:  =  16. 


144  ALGEBRA. 

22.  A  is  72  years  old,  and  B's  age  is  two-thirds  of  A's. 

How  long  is  it  since  A  was  five  times  as  old  as  B  ? 

23.  A  mother  is  70  years  old,  her  daughter  is  half  that 

age.     How  long  is  it  since  the  mother  was  three 
and  one-third  times  as  old  as  the  daughter  ? 

24.  A  father  is  three  times  as  old  as  the  son ;  four  years 

ago  the  father  was  four  times  as  old  as  the  son  then 
was.     What  is  the  age  of  each  ? 

25.  A  is  twice  as  old  as  B,  and  seven  years  ago  their 

united   ages  amounted  to  as  many  years  as  now 
represent  the  age  of  A.     Find  the  ages  of  A  and  B. 

26.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  25  years  ;  the  difference  is  one-third  what 

the  sum  will  be  in  20  years.  What  is  the  age  of  each  ? 

Note.     If  A  can  do  a  piece  of  work  in  x  days,  the  'part  of  the 

work  that  he  can  do  in  one  day  will  be  represented  by  -.     Thus,  if 

he  can  do  the  work  in  5  days,  in  1  day  he  can  do  \  of  the  work. 

Ex.    A  can  do  a  piece  of  work  in  5  days,  and  B  can  do  it 

in  4  days.    How  long  will  it  take  A  and  B  together  ? 

Let        X  =  the  number  of  days  it  will  take  A  and  B  together. 
Then      -;  =  the  part  they  can  do  in  one  day. 
Now,     \  =  the  part  A  can  do  in  one  day, 
and  \  =  the  part  B  can  do  in  one  day. 

•*•  i  +  4  =  the  part  A  and  B  can  do  in  one  day. 

4a;  +  5a;  =  20, 
9a;  =  20, 

a;  =  2f. 

Therefore  they  will  do  the  work  in  2f  days. 

27.  A  can  do  a  piece  of  work  in  5  days,  B  in  6  days,  and 

C  in  71  days;    in  what  time  will  they  do  it,  all 
working  together  ? 


PROBLEMS.  145 

28.  A  can  do  a  piece  of  work  in  2^  days,  B  in  3J  days, 

and  0  in  3|  days ;  in  what  time  will  they  do  it,  all 
working  together  ? 

29.  Two  men  who  can  separately  do  a  piece  of  work  in 

15  days  and  16  days,  can,  with  the  help  of  another, 
do  it  in  6  days.  How  long  would  it  take  the  third 
man  to  do  it  alone  ? 

30.  A  can  do  half  as  much  work  as  B,  B  can  do  half  as 

much  as  C,  and  together  they  can  complete  a  piece 
of  work  in  24  days.  In  what  time  can  each  alone 
complete  the  work  ? 

31.  A  does  f  of  a  piece  of  work  in  10  days,  when  B  comes 

to  help  him,  and  they  finish  the  work  in  3  days 
more.  How  long  would  it  have  taken  B  alone  to 
do  the  whole  work  ? 

32.  A  and  B  together  can  reap  a  field  in  12  hours,  A  and 

C  in  16  hours,  and  A  by  himself  in  20  hours.  In 
what  time  can  B  and  C  together  reap  it  ?  In  what 
time  can  A,  B,  and  0  together  reap  it  ? 

33.  A  and  B  together  catti  do  a  piece  of  work  in  12  days, 

A  and  0  in  15  days,  B  and  C  in  20  days.  In  what 
time  can  they  do  it,  all  working  together? 

Note.     If  a  pipe  can  fill  a  vessel  in  x  hours,  the  part  of  the 
vessel  filled  by  it  in  one  hour  will  be  represented  by  -.     Thus,  if  a 

X 

pipe  will  fill  a  vessel  in  3  hours,  in  1  hour  it  will  fill  |  of  the  vessel. 

34.  A  tank  can  be  filled  by  two  pipes  in  24  minutes  and 

30  minutes  respectively,  and  emptied  by  a  third  in 
20  minutes.  In  what  time  will  it  be  filled  if  all 
three  are  running  together  ? 

35.  A  tank  can  be  filled  in  15  minutes  by  two  pipes,  A 

and  B,  running  together.     After  A  has  been  run- 


146  ALGEBRA. 

ning  by  itself  for  5  minutes,  B  is  also  turned  on, 
and  the  tank  is  filled  in  13  minutes  more.  In  what 
time  may  it  be  filled  by  each  pipe  separately  ? 

36.  A  cistern  could  be  filled  by  two  pipes  in  6  hours  and 

8  hours  respectively,  and  could  be  emptied  by  a 
third  in  12  hours.  In  what  time  would  the  cistern 
be  filled  if  the  pipes  were  all  running  together  ? 

37.  A  tank  can  be  filled  by  three  pipes  in  1  hour  and  20 

minutes,  3  hours  and  20  minutes,  and  5  hours,  re- 
spectively. In  what  time  will  the  tank  be  filled 
when  all  three  pipes  are  running  together  ? 

38.  If  three  pipes  can  fill  a  cistern  in  a,  5,  and  c  minutes, 

respectively,  in  what  time  will  it  be  filled  by  all 
three  running  together  ? 

39.  The  capacity  of  a  cistern  is  755i  gallons.     The  cistern 

has  three  pipes,  of  which  the  first  lets  in  12  gallons 
in  3i  minutes,  the  second  15i  gallons  in  2\  minutes, 
the  third  17  gallons  in  3  minutes.  In  what  time 
will  the  cistern  be  filled  by  the  three  pipes  running 
together  ? 

Note.     In  questions  involving  distance,  time,  and  rate, 

Distance      m- 

=  iime. 

Rate 

Thus,  if  a  man  travels  40  miles  at  the  rate  of  4  miles  an  hour, 
—  ==  number  of  hours  required. 

Ex.  A  courier  who  goes  at  the  rate  of  31^  miles  in  5  hours, 
is  followed,  after  8  hours,  by  another,  who  goes  at 
the  rate  of  22^  miles  in  3  hours.  In  how  many 
hours  will  the  second  overtake  the  first? 

Since  the  first  goes  31^  miles  in  5  hours,  his  rate  per  hour  is  6^3^ 
miles. 


PROBLEMS.  147 

Since  the  second  goes  22.}  miles  in  3  hours,  his  rate  per  hour  is  7j 
miles. 

Let  X  =  the  number  of  hours  the  first  is  travelling. 

Then  a;  —  8  =  the  number  of  hours  the  second  is  travelling. 

Then  6y*o '^  =  t^®  ^^'^b^^  <^^  "liles  the  first  travels  ; 

{x  —  8)  7^  =  the  number  of  miles  the  second  travels. 
They  both  travel  the  same  distance. 

.-.  Q^\x^{x-%)11 
The  solution  of  which  gives  42  hours. 

40.  A  sets  out  and  travels  at  the  rate  of  7  miles  in  5  hours. 

Eight  hours  afterwards,  B  sets  out  from  the  same 
place  and  travels  in  the  same  direction,  at  the  rate 
of  5  miles  in  3  hours.  In  how  many  hours  will  B 
overtake  A? 

41.  A  person  walks  to  the  top  of  a  mountain  at  the  rate 

of  2i  miles  an  hour,  and  down  the  same  way  at  the 
rate  of  3 2  miles  an  hour,  and  is  out  5  hours.  How 
far  is  it  to  the  top  of  the  mountain  ? 

42.  A  person  has  a  hours  at  his  disposal.     How  far  may 

he  ride  in  a  coach  which  travels  b  miles  an  hour,  so 
as  to  return  home  in  time,  walking  back  at  the  rate 
of  c  miles  an  hour  ? 

43.  The  distance  between  London  and  Edinburgh  is  360 

miles.  One  traveller  starts  from  Edinburgh  and 
travels  at  the  rate  of  10  miles  an  hour ;  another 
starts  at  the  same  time  from  London,  and  travels  at 
the  rate  of  8  miles  an  hour.  How  far  from  London 
will  they  meet  ? 

44.  Two  persons  set  out  from  the  same  place  in  opposite 

directions.  The  rate  of  one  of  them  per  hour  is  a 
mile  less  than  double  that  of  the  other,  and  in  4 
hours  they  are  32  miles  apart.  Determine  their 
rates. 


148  ALGEBRA. 

45.  In  going  a  certain  distance,  a  train  travelling  35  miles 

an  hour  takes  2  hours  less  than  one  travelling  25 
miles  an  hour.     Determine  the  distance. 

Note.     In  problems  relating  to  clocks,  it  is  to  be  noticed  that  the 
minute-hand  moves  twelve  times  as  fast  as  the  hour-hand. 

Ex.    Find  the  time   between  2  and  3  o'clock  when  the 
hands  of  a  clock  are  together. 

At  2  o'clock  the  hour-hand  is  10  minute-spaces  ahead  of  the 
minute-hand. 

Let  X  =  the  number  of  spaces  the  minute-hand  moves  over. 

Then  ar  —  10  =  the  number  of  spaces  the  hour-hand  moves  over. 
Now,  as  the  minute-hand  moves  12  times  as  fast  as  the  hour-hand, 
12(ic  — 10)=  the  number  of  spaces  the  minute-hand  moves  over. 
.-.  a;  =  12  (a;  -  10), 
and  11  a;  =  120. 

.-.  X  =  lO^f 

Therefore  the  time  is  10^  minutes  past  2  o'clock. 

46.  At  what  time  are  the  hands  of  a  watch  together  : 

I.    Between  3  and  4  ? 

II.    Between  6  and  7  ? 

III.   Between  9  and  10? 

47.  At  what  time  are  the  hands  of  a  watch  at  right  angles : 

I.    Between  3  and  4  ? 

II.    Between  4  and  5  ? 

III.    Between  7  and  8? 

48.  At  what  time  are  the  hands  of  a  watch  opposite  to 

each  other : 

I.    Between  1  and  2? 

II.    Between  4  and  5  ? 

III.    Between  8  and  9? 


PROBLEMS.  149 

49.  It  is  between  2  and  3  o'clock;  but  a  person  looking  at 

his  watch  and  mistaking  the  hour-hand  for  the 
minute-hand,  fancies  that  the  time  of  day  is  55 
minutes  earlier  than  it  really  is.  What  is  the  true 
time? 

Note.  If  a  represents  the  number  of  feet  in  the  length  of  a  step 
or  leap,  and  x  the  number  of  steps  or  leaps  taken,  then  ax  will  repre- 
sent the  number  of  feet  in  the  distance  made, 

Ex.  A  hare  takes  4  leaps  to  a  greyhound's  3 ;  but  2  of  the 
greyhound's  leaps  are  equivalent  to  3  of  the  hare's. 
The  hare  has  a  start  of  50  leaps.  How  many  leaps 
must  the  greyhound  take  to  catch  the  hare  ? 

Let    3  a;  =  the  number  of  leaps  taken  by  the  greyhound. 
Then  4  a;  =  the  number  of  leaps  of  the  hare  in  the  same  time. 
Also,  let  a  denote  the  number  of  feet  in  one  leap  of  the  hare. 

Then  —  will  denote  the  number  of  feet  in  one  leap  of  the 
greyhound. 

That  is,         3  X  X  —  =  the  whole  distance, 

A 

and  (50  +  4x)  a  =  the  whole  distance. 

.  —  =  (50  + 4a;)  a. 

Divide  by  «,  —  =  50+4x, 

9a;  =100 +  8  a;, 
X  -  100. 
.^  3  a;  =  300. 

Thus  the  greyhound  must  take  300  leaps. 

50.  A  hare  takes  6  leaps  to  a  dog's  5,  and  7  of  the  dog's 

leaps  are  equivalent  to  9  of  the  hare's.  The  hare 
has  a  start  of  50  of  her  own  leaps.  How  many  leaps 
will  the  hare  take  before  she  is  caught? 


150  ALGEBRA. 

51.  A  greyhound  makes  4  leaps  while  a  hare  makes  5 ;  but 

3  of  the  greyhound's  leaps  are  equivalent  to  4  of  the 
hare's.  The  hare  has  a  start  of  60  of  the  greyhound's 
leaps.  How  many  leaps  does  each  take  before  the 
hare  is  caught? 

52.  A  greyhound  makes  two  leaps  while  a  hare  makes  3  ; 

but  1  leap  of  the  greyhound  is  equivalent  to  2  of  the 
hare's.  The  hare  has  a  start  of  80  of  her  own  leaps. 
How  many  leaps  will  the  hare  take  before  she  is 
caught  ? 

Note.  If  the  number  of  units  in  the  breadth  and  length  of  a 
rectangle  is  represented  by  x  and  x  -{-  a,  respectively,  then  x{x  +  a) 
will  represent  the  number  of  units  of  area  in  the  rectangle,  the  unit 
of  area  having  the  same  name  as  the  linear  unit  in  which  the  sides 
of  the  rectangle  are  expressed. 

53.  A  rectangle  whose  length  is  5  feet  more  than  its  breadth 

would  have  its  area  increased  by  22  feet  if  its  length 
and  breadth  were  each  made  a  foot  more.  Find  its 
dimensions. 

54.  A  rectangle  has  its  length  and  breadth  respectively  5 

feet  longer  and  3  feet  shorter  than  the  side  of  the 
equivalent  square.     Find  its  area. 

55.  The  length  of  a  rectangle  is  an  inch  less  than  double 

its  breadth ;  and  when  a  strip  3  inches  wide  is  cut 
off  all  round,  the  area  is  diminished  by  210  inches. 
Find  the  size  of  the  rectangle  at  first. 

56.  The  length  of  a  floor  exceeds  the  breadth  by  4  feet ;  if 

each  dimension  were  increased  by  1  foot,  the  area 
of  the  room  would  be  increased  by  27  square  feet. 
Find  its  dimensions. 

Note.    If  b  pounds  of  metal  lose  a  pounds  when  weighed  in  water, 

1  pound  will  lose  -  of  a  pounds,  or  ^  of  a  pound. 
b  b 


PROBLEMS.  151 

57.  A  mass  of  tin  and  lead  weighing  180  pounds  loses  21 

pounds  when  weighed  in  water;  and  it  is  known  that 
37  pounds  of  tin  lose  5  pounds,  and  23  pounds  of 
lead  lose  2  pounds,  when  weighed  in  water.  How 
many  pounds  of  tin  and  of  lead  in  the  mass  ? 

58.  If  19  pounds  of  gold  lose  1  pound,  and  10  pounds  of 

silver  lose  1  pound,  when  weighed  in  water,  find  the 
amount  of  each  in  a  mass  of  gold  and  silver  weigh- 
ing 106  pounds  in  air  and  99  pounds  in  water. 

59.  Fifteen  sovereigns  should  weigh  77  pennyweights  ;  but 

a  parcel  of  light  sovereigns,  having  been  weighed  and 
counted,  was  found  to  contain  9  more  than  was  sup- 
posed from  the  weight ;  and  it  appeared  that  21  of 
these  coins  weighed  the  same  as  20  true  sovereigns. 
How  many  were  there  in  all  ? 


60.  There  are  two  silver  cups,  and  one  cover  for  both.   The 

first  weighs  12  ounces,  and  with  the  cover  weighs 
twice  as  much  as  the  other  without  it ;  but  the  sec- 
ond with  the  cover  weighs  one-third  more  than  the 
first  without  it.     Find  the  weight  of  the  cover. 

61.  A  man  wishes  to  inclose  a  circular  piece  of  ground  with 

palisades,  and  finds  that  if  he  sets  them  a  foot  apart 
he  will  have  too  few  by  150 ;  but  if  he  sets  them  a 
yard  apart  he  will  have  too  many  by  70.  What  is 
the  circuit  of  the  piece  of  ground  ? 

62.  A  horse  was  sold  at  a  loss  for  $200;  but  if  it  had  been 

sold  for  $250,  the  gain  would  have  been  three-fourths 
of  the  loss  when  sold  for  $200.  Find  the  value  of 
the  horse. 

63.  A  and  B  shoot  by  turns  at  a  target.     A  puts  7  bullets 

out  of  12,  and  B  9  out  of  12,  into  the  centre.  Be- 
tween them  they  put  in  32  bullets.  How  many 
shots  did  each  fire  ? 


152  ALGEBRA. 

64.  A  boy  buys  a  number  of  apples  at  the  rate  of  5  for  2 

pence.  He  sells  half  of  them  at  2  a  penny  and  the 
rest  at  3  a  penny,  and  clears  a  penny  by  the  trans- 
action.    How  many  does  he  buy  ? 

65.  A  person  bought  a  piece  of  land  for  $6750,  of  which 

he  kept  f  for  himself.  At  the  cost  of  $260  he  made 
a  road  which  took  y^-g-  of  the  remainder,  and  then  sold 
the  rest  at  12i  cents  a  square  yard  more  than  double 
the  price  it  cost  him,  thus  clearing  his  outlay  and 
$500  besides.  How  much  land  did  he  buy,  and 
what  was  the  cost-price  per  yard  ? 

66.  A  boy  who  runs  at  the  rate  of  12  yards  per  second 

starts  20  yards  behind  another  whose  rate  is  10^ 
yards  per  second.  How  soon  will  the  first  boy  be 
10  yards  ahead  of  the  second  ? 

67.  A  merchant  adds  yearly  to  his  capital  one-third  of  it, 

but  takes  from  it,  at  the  end  of  each  year,  $5000  for 
expenses.  At  the  end  of  the  third  year,  after  de- 
ducting the  last  $5000,  he  has  twice  his  original 
capital.     How  much  had  he  at  first  ? 

68.  A  shepherd  lost  a  number  of  sheep  equal  to  one-fourth 

of  his  flock  and  one-fourth  of  a  sheep ;  then,  he  lost 
a  number  equal  to  one-third  of  what  he  had  left  and 
one-third  of  a  sheep ;  finally,  he  lost  a  number  equal 
to  one-half  of  what  now  remained  and  one-half  a 
sheep,  after  which  he  had  but  25  sheep  left.  How 
many  had  he  at  first  ? 

69.  A  trader  maintained  himself  for  three  years  at  an  ex- 

pense of  $250  a  year ;  and  each  year  increased  that 
part  of  his  stock  which  was  not  so  expended  by  one- 
third  of  it.  At  the  end  of  the  third  year  his  original 
stock  was  doubled.     What  was  his  original  stock  ? 


PROBLEMS.  153 

70.  A  cask  contains  12  gallons  of  wine  and  18  gallons  of 

water ;  another  cask  contains  9  gallons  of  wine  and 
3  gallons  of  water.  How  many  gallons  must  be 
drawn  from  each  cask  to  produce  a  mixture  contain- 
ing 7  gallons  of  wine  and  7  gallons  of  water? 

71.  The  members  of  a  club  subscribe  each  as  many  dollars 

as  there  are  members.  If  there  had  been  12  more 
members,  the  subscription  from  each  would  have 
been  $10  less,  to  amount  to  the  same  sum.  How 
many  members  were  there  ? 

72.  A  number  of  troops  being  formed  into  a  solid  square, 

it  was  found  there  were  60  men  over;  but  when 
formed  in  a  column  with  5  men  more  in  front  than 
before,  and  3  men  less  in  depth,  there  was  lacking 
one  man  to  complete  it.    Find  the  number  of  troops. 

73.  An  officer  can  form  the  men  of  his  regiment  into  a 

hollow- square  twelve  deep.  The  number  of  men  in 
the  regiment  is  1296.  Find  the  number  of  men  in 
the  front  of  the  hollow  square. 

74.  A  person  starts  from  P  and  walks  towards  Q  at  the 

rate  of  3  miles  an  hour ;  20  minutes  later  another 
person  starts  from  Q  and  walks  towards  P  at  the 
rate  of  4  miles  an  hour.  The  distance  from  P  to  Q 
is  20  miles.     How  far  from  P  will  they  meet  ? 

75.  A  person  engaged  to  work  a  days  on  these  conditions: 

for  each  day  he  worked  he  was  to  receive  b  cents, 
and  for  each  day  he  was  idle  he  was  to  forfeit  c 
cents.  At  the  end  of  a  days  he  received  d  cents. 
How  many  days  was  he  idle  ? 

76.  A  banker  has  two  kinds  of  coins :  it  takes  a  pieces  of 

the  first  to  make  a  dollar,  and  b  pieces  of  the  second 
to  make  a  dollar.  A  person  wishes  to  obtain  c  pieces 
for  a  dollar.  How  many  pieces  of  each  kind  must 
the  banker  give  him  ? 


CHAPTER  XL 

SIMULTANEOUS   EQUATIONS   OF   THE   FIRST 
DEGREE. 

155.  If  we  have  two  unknown  numbers  and  but  one  rela- 
tion between  them,  we  can  find  an  unlimited  number  of 
pairs  of  values  for  which  the  given  relation  will  hold  true. 
Thus,  if  X  and  y  are  unknown,  and  we  have  given  only  the 
one  relation  x  -\-y  =  10,  we  can  assume  any  value  for  x, 
and  then  from  the  relation  x-{-  y  =10  find  the  correspond- 
ing value  of  y.  For  from  :r  +  y  =  10  we  find  y  =  10  —  x. 
If  X  stands  for  1,  y  stands  for  9;  if  :r  stands  for  2,  y  stands 
for  8 ;  if  a;  stands  for  —  2,y  stands  for  12  ;  and  so  on  with- 
out end. 

156.  We  may,  however,  have  two  equations  that  express 
different  relations  between  the  two  unknowns.  Such  equa- 
tions are  called  independent  equations.  Thus,  x-\-y  =10 
and  X  —  y  =  2  are  independent  equations,  for  they  evidently 
express  different  relations  between  x  and  y. 

157.  Independent  equations  involving  the  same  unknowns 
are  called  simultaneous  equations. 

If  we  have  two  unknowns,  and  have  given  two  indepen- 
dent equations  involving  them,  there  is  but  one  pair  of  values 
which  will  hold  true  for  both  equations.  Thus,  if  in  §  156, 
besides  the  relation  a;  -f  y  =  10,  we  have  also  the  relation 
a;  -T  y  =  2,  the  only  pair  of  values  for  which  both  equations 
will  hold  true  is  the  pair  a:  =  6,  y  =  4. 

Observe  that  in  this  problem  x  stands  for  the  same  num- 
ber in  both  equations  ;  so  also  does  y. 


SIMULTANEOUS   EQUATIONS.  155 

158.  Simultaneous  equations  are  solved  by  combining 
the  equations  so  as  to  obtain  a  single  equation  with  one 
unknown  number ;  this  process  is  called  elimination. 

There  are  three  methods  of  elimination  in  general  use : 

I.   By  Addition  or  Subtraction. 
II.    By  Substitution. 
III.    By  Comparison. 

159.  Elimination  by  Addition  or  Subtraction. 


(1)   Solve:              5a;-3y  =  20| 
2:r+5y  =  39i 

(1) 
(2) 

Multiply  (1)  by  5,  and  (2)  by  3, 

25x-15i/  =  100 
6a; +  152/ =117 

(3) 
(4) 

Add  (3)  and  (4),          31  x             =217 

.-.  X  =  7. 
Substitute  the  value  of  x  in  (2), 

14 +  52/ =  39. 
.-.2/ =  5. 

In  this  solution  y  is  eliminated  by  addition. 

(2)   Solve:            6a;  +  352/ =  177  ) 
^x-2\y=    33j 

(1) 
(2) 

Multiply  (1)  by  4,  and  (2)  by  3. 

24x  + 1403/ =  708 
24a;-    633/=    99 

(3) 
(4) 

Subtract,  203  y  =  609 

••■y  =  3. 

Substitute  the  value  of  y  in  (2), 

8x-63  =  33. 
.-.  a;  =  12. 

In  this  solution  x  is  eliminated  by  subtraction. 


156  ALGEBRA. 

160.  Hence,  to  eliminate  by  addition  or  subtraction,  we 
have  the  following  rule  : 

Multiply  the  equations  hy  such  numbers  as  will  make  the 
coefficients  of  one  of  the  unknown  number's  equal  in  the 
resulting  equations. 

Add  the  resulting  equations,  or  subtract  one  from  the  other, 
according  as  these  equal  coefficients  have  unlike  or  like  signs. 

Note,  It  is  generally  best  to  select  the  letter  to  be  eliminated 
which  requires  the  smallest  multipliers  to  make  its  coelBficients  equal ; 
and  the  smallest  multiplier  for  each  equation  is  found  by  dividing 
the  L.  C.  M.  of  the  coefficients  of  this  letter  by  the  given  coefficient  in 
that  equation.  Thus,  in  example  (2),  the  L.  C.  M.  of  6  and  8  (the  co- 
efficients of  x)  is  24,  and  hence  the  smallest  multipliers  of  the  two 
equations  are  4  and  3  respectively. 

Sometimes  the  solution  is  simplified  by  first  adding  the 
given  equations,  or  by  subtracting  one  from  the  other. 

(3)  a:  +  492/=    51  (1) 

49a;  +      3/ -    99  (2) 

Add  (1)  and  (2),  50a;  +  50y  =  150  (3) 

Divide  (3)  by  50,  k  +  y  =  3.  (4) 

Subtract  (4)  from  (1),  48^  =  48. 

••■2/  =  l- 
Subtract  (4)  from  (2),  48  a;  -  96. 

.-.  a;  =  2. 


Exercise  62. 
Solve  by  addition  or  subtraction : 

2.      x—2y  =  4:')    4.    3:^-5y  =  51|  6.    3a;+2y-=39) 

2x-    y  =  b)          2a;  +  7?/=   3)  3y-2a;  =  13j 


SIMULTANEOUS   EQUATIONS.  167 

7.  3.r-4y  =  -5) 
4:r  — 5y=      1) 

8.  ll:r  +  3y=100) 

Ax—7y=     43 

9.  a:  +  49y  =  693) 
49a;  +      y  =  357  j 

10.    17a;  +  32/  =  57)  14.    17:r  +  30y  =  59) 

16y  -  3:u  =  23  J  19a;  +  28y  =  77  j 


11. 

12a; 

+    73/ 

=  176) 
=      3j 

3y 

-19a; 

12. 

2a;- 
4y- 

-7y  = 
-9a;  = 

^1 
19] 

13. 

69y 
14a; 

-17a; 
-13y 

=    103) 
=  -41 1 

161.   Elimination  by  Substitution. 

(1)   Solve:  5a;  +  4y  =  32 

4a;  +  3y  =  25 


5.'c  +  4y  =  32. 

(1) 

4a; +  83/ =  25. 

(2) 

Transpose  4 2/ in  (1),                  5x  =  32-4y. 

(3) 

Divide  by  coefficient  of  x,            x  -  ^^"^.V- 

(4) 

Substitute  the  value  of  x  in  (2), 

4(?2^).3,.., 

l^»-l«y+3y      25, 
5 

128  -  16y  +  157/  =  125, 

-3/  =  -3. 

••.y  =  3. 

Substitute  the  value  of  y  in  (2), 

4a; +  9  =  25. 

.-.  X  =  4. 

Hence,  to  eliminate  by  substitution, 

From  one  of  the  equations  obtain  the  value  of  one  of  the 
unknown  numbers  in  terms  of  the  other. 

Substitute  for  this  unknown  number  its  value  in  the  other 
equation,  and  reduce  the  resulting  equation. 


158  ALGEBRA. 

Exercise  63. 
Solve  by  substitution : 
1.    Sx  —  4:^  =  2\  8.    Sx-4:y 


7)  Sx  +  2y=    0) 


1x-9y=1)  Sx  +  2y 

2.  1x-by=^24:^  9.    9a; -5?/ =  52  1 
4a;-3y=ll3  8i/-Sx=    8) 

3.  3a;  +  2y  =  32j  10.      5a;-3y=    4| 
20a;-3y=    13  12y-7a;  =  10/ 

4.  ll:i;-7y  =  37)  11.      9y-7a;=13) 

8:r  +  9y  =  41j  15:r-7y=   9  3. 

5.  lx-\-    5y=60)  12.      5a;-2y=    51) 
13:r-lly=10  3  19:r-3y  =  180  3 

6.  6a;- 7?/ =  42 1  13.    4a;  +    9y=106| 
=  75i 


7a;  -  6y  =  75  )  8a;  +  17y  =  198  3 

10a;+    9y  =  290|  14.      8a;  +  3y  =  3 

12a;-lly  =  130  3  12a;  +  9y  =  3 


162.   Elimination  by  Comparison. 

Solve:  2a;-5y  =  66) 

3a;  +  2y  =  23  3 

2x-5y  =  66.  (1) 

3  a; +  23/ =  23.  (2) 
Transpose  5  3/  in  (1),  and  2y  m  (2), 

2a;  =  66  4  by,  (3) 

3a;  =  23-2y.  (4) 

Divide  (3)  by  2,  x^  66j^  ^^^ 


66  +  5i 

2 
23 -2i 


Divide  (4)  by  3,  a;  =  ±i^-^.  (6) 

Equate  the  values  of  x,  §i±li/  =  23-_2^_  ^^^ 

}2  o 


SIMULTANEOUS 

EQUATIONS. 

Reduce  (7), 

198  +  152/  = 

46 -4y, 

192/ = 

:  -  152. 

'■y  = 

:-8. 

Substitute  the  value  of  y  in  (1), 

2x  +  40  = 

■m. 

.•.  x=- 

.13. 

159 


163.    Hence,  to  eliminate  by  comparison, 

From  each  equation  obtain  the  value  of  one  of  the  unknown 
numbers  in  terms  of  the  other. 

Form  an  equation  from  these  equal  values  and  reduce  the 
equation. 

Exercise  64. 

Solve  by  comparison  : 

1.  a:+15y  =  53)  8.    3y-7a:=    4| 

2.  4:x+    92/ =  511  9.    21y  +  20:i;=165") 
8a:-13y=    9  I  772/ -  30a;  =  295  j 

3.  4a;  +  3y==48|  10.    lla:-10y=14| 
5y-3a;  =  22j  5:r+    7y-4lj 

4.  2x  +  Sy  =  iS)  11.    72/ -3.1;  =  139) 
10^-    2/-    73  2a;  +  5y=    91) 

5.  6x~    1y=    33|  12.    llx+\2y=    69) 
1U+122/  =  100J  19a;-   42/=1533 

6.  5a;+72/  =  43)  13.    24:r+    7y=    27) 
llo;  +  9y  =  69  3  8a;  -  332/  =  115  3 

7.  8.T-2l2/==    33)  14.   a;  =  By -19) 
6a;  +  35y  =  177  3  y  =  3a;  -  23  3 


160  ALGEBRA. 

164.    Each   equation   must   be   simplified,  if  necessary, 
before  the  elimination. 

Solve:  H-^(7/-{-l)  =  ll 

i(a;+l)  +  f(2/-l)  =  9j 

ia;~i(2/  +  l)  =  l.  (1) 

i(a^  +  l)  +  f(2/-l)  =  9.  (2) 

Multiply  (1)  by  4,  and  (2)  by  12, 

3a;-23/-2  =  4.  (3) 

4a: +  4 +  97/ -9  =  108.  (4) 

From  (3),  3x-2y^6.  ^         (5) 

From  (4),  4x  +  dy  =  m.  (6) 

Multiply  (5)  by  4,  and  (6)  by  3, 

12x-    8y=    24 
12a; +  272/ =  339 

353/ =  315 
.-.2/ =  9. 
Substitute  value  of  3/  in  (1),  a;  =  8. 

Exercise  65. 

Solve : 


2 


5  5  10 


y-r      g  .  6  4 

5.    (^+l)(y  +  2)-(:r+2)(2/+l)  =  -l| 


3(x  +  3)-4(y  +  4)=- 

1 


.r-2      10-a;_y-10 
5  3  4 

3  8  4      J 


SIMULTANEOUS   EQUATIONS. 


161 


3  4  5 


x-S      y-3 


2y  —  X 


15    ^z:4_2Mi2] 

5  10 

^  ,  y  -  ^  -  3 

6^     4 


3r-2y  5a:-3y^^     j^ 

5  3 

2:r-3.y  4a;-3y_        ^ 

3  2  ^ 


16.  ^^i-12y^9 

l-3a;_ll-3y 

7  5 


**  3  4 

3a;-4y  +  3  ,  4a;-2y-9_. 
4  3^         ■ 


10.    11.;  =  42/ +  4^1  17.    5a;-i(5y  +  2)-32| 

3y  +  i(^  +  2)==9      J 


11, 


4i^  =  i2/-21A) 
13  3 


:r  +  2y+3  4:X-by-{-6 

3  _         19 

6a;-5y  +  4    '  Sx+2i/-]-l 


18.    3rr-0.252/-28     | 
0.12a;+0.72/=2.54( 


12.    ^±^  =  15 

y-^     8 

g^_3y  +  44^^QQ 


19.    7(0;- 1)- 3(3/  + 8) 

4^'  +  2^5y  +  9 

9  2 


13     3a:-5y  .  3^2a;+y 
2  5 

Q      a;-2y„a;     y 
4  2^3 


20.    7a;+i(2y  +  4)=16) 
3y-i(:r+2)=8      i 


14     4a;-3y-7_3a;     2y     5 
5  10      15      6 

^lJL^_3^_-^_^^^  ,  ^  ,  i_ 
3     ^2     20  15    ^6^  10  J 


162 


ALGEBRA. 


21.    5^^4-3a;  =  4y-2 

5rr  +  6y      3a;-2y_o        o 
6  4*^ 


22     5a;-3      3a;-19_^      ^y  -  x 

2  2  3 

2rg  +  y  _  9:g-7  ^.  30/ +  3)  __  4a.-+5y 
2  8*4  16 

(^+7)(y-2)+3  =  2:ry-(y-a)(.:+l)J 


6a;  +  9      3:r  +  5y_3,      3:r  +  4 
4      "^  4a:-6  "^      2 

8y+7      6:r-3.y_,      4y-9 
10  2y-8  "^      5 


25.    ^_g£ii5^20-^^-^^ 
23 -a;  2 

y^l3_^30_73-3.y 


2/4- 


.r-18 


165.   Literal  Simultaneous  Equations. 

Solve  :  ax-\-hy  =  c 

a^x  -\-  Vy  —  c 


.} 


Note,  The  letters  a^,  b^  are  read  a  prime,  b  prime.  In  like  man- 
ner, a^^,  a'^f  are  read  a  second,  a  third,  and  a^,  a^,  a^  are  read  a  sub 
one,  a  sub  two,  a  sub  three.  It  is  sometimes  convenient  to  represent 
different  numbers  that  have  a  common  property  by  the  same  letter 
marked  by  accents  or  suffixes.  Here  a  and  a^  have  a  common  prop- 
erty as  coefficients  of  x. 

ax  +  by  =  c.  (1) 


a^x  -t-  yy  =  c^. 


(2) 


SIMULTANEOUS   EQUATIONS.  163 


To  find  the  value  of  y,  multiply  (1)  by  a\  and  (2)  by  a, 
aa^x  +  a^hy  =  a^c 


a^by  —  ab^y  =  a'c  —  ac^ 
a'c  —  ac^ 


y 


a'b  -  ab^ 


To  find  the  value  of  x,  multiply  (1)  by  b\  and  (2)  by  6,  and  pro- 
ceed as  in  finding  the  value  of  y. 


Exercise  66. 


Solve : 

1.  x-{-y  —  a\  3.    mx -\-ny  =  a)    5.     mx  —  ny^=r  ) 
x  —  y=h)  px  -^  qy  =  b  )  m'x  -{-  n'y  =  r'  ) 

2.  ax-\-by  =  c\  4.    ax -{- by  =  e  }      6.    ax-\-by  =  c) 
px-\-qy  =  r  )  ax -{- cy  —  d )  dx -{- fy  ■=  c^  ) 

7.    5  +  ^=c      1  12.    ^^^Zl+1  = 

a      b  1  x—y— 1 


b     a  X  ■\-y  —  \ 

8.    abx  +  cdy  =  2^  ^^     ax^by-^'^^^ 


d-b \  "   '       2 

ax  —  cy  —  —-—-  /        7N         , 

bd    )  {a  —  o)x  =  {a 

a     _      b      ^  14.    ax-\-by  —  c^ 


\ 
+1-).'/) 


b-\-y      2>a-\-x 


ax-\-2by  =  d    J  b  -\-y     a-\- x 


0 


10.  -^ ^  =  -m  15.    -^  +  -^^2a 

a  +  6      a  —  o      a-\-o  I  a  +  o      a  — 6 

X      .      y     __     \       I  x  —  y_  x-\-y 

a  +  b      a-b'~~a-b  )  2ab  ~~ a^-^h" 

11.  a(a  — rp)  =  5(a;+2/— a)  )  16.    bx  —  bc  =  ay  —  ac}^ 
a(y  —  b  —  x)  =  b(y  —  b))  x  —  y  =  a  —  b        ) 


164  ALGEBRA. 

17.  ^^=c 

y-  b 

a{x  —  a)  -{-  b  {y  —  h)  -\-  ahc  —  0 

18.  {a-\-h)x-{a-h)y  =  ^ab  ") 


{a -  h)x -\-  {a-]-h)y  =  2a^ -21/ 

{x-\-a){y-{-h)-{x  —  i 
x-y-{-2(a-h)  =  {) 

(a  +  b){x  +  3/)  -  (a  - 
{a-h)(x  +  y)  +  {a-\-b){x-y) 


19.  {x-\-a){y-\-b)-{x-a){y-b)  =  2{a-b)n 

20.  {a  +  b){x^y)~{a-b){x-y)  =  o}\ 

b'] 


166.  Fractional  simultaneous  equations,  of  which  the  de- 
nominators are  simple  expressions  and  contain  the  unknown 
numbers,  may  be  solved  as  follows : 


(1)  Solve : 

a 

X 

,  ^       1 
y 
,d 

c 

X 

y       J 

-  +  -  =  m. 

X     y 

X      y 

(1) 
(2) 

Multiply  (1)  by  c, 

ac  .  he 
X       y 

(3) 

Multiply  (2)  by  a, 

ac  ,  ad 

_  +  _  =  an. 

X       y 

(4) 

Subtract  (4)  from  (3), 

he  —  ad 

-=cm- 

y 

he  —  ad  =  {cm  ■ 

-  an. 

Multiply  both  sides  by 

y. 

-  «^)y- 

cm  - 

-ad 

-  an 

Multiply  (1)  by  d, 

X       y 

(5) 

Multiply  (2)  by  6, 

k  +  M^in. 

(6) 

SIMULTANEOUS   EQUATIONS. 


165 


Subtract  (6)  from  (5),                ^f:^f:_j^  _  ^^ 
Multiply  both  sides  by  x,         ad-bc  =  {dm 

dm 

-6n. 

-6n)a;. 

-6c 

-bn 

(2)  Solve: 

A+A=7l 

Bx     by 

7-      1   =3 

ax     lOy         J 

' 

We  have                                  A  +  A  =  7, 
3a;     5y 

and                                               ^_J_  =  3. 

6a;      lOy 

Multiply  (1)  by  15,  the  L.  C.  M.  of  3  and  5,  and  (2)  by 

2^  +  6^105. 
X      y 

30, 

(1) 

(2) 
(3) 

X      y 
Multiply  (4)  by  2,  and  add  the  result  to  (3), 

?5  =  285 

X 

1 

••••"  =  3 

(4) 

Substitute  the  value  of  x  in  (1),  and  we  get 

y-. 

Solve  : 

Exercise  67. 

1.   1  +  2=10 
X     y 

^  +  5^20 
X     y 

1      3   2_A-Ai 

'    a;     Sy      27 
1    ,  1_11 

J              ^x~^y'    72  J 

5. 

3_ 

X 

4_ 
a; 

4^ 
5^ 

y 

:5" 

=  6 

2.    1  +  2  =  „1 
X     y 

§  +  4  =  6 

X     y         J 

4.    1  +  §  =  41 
X     y 

3_2^4 

>. 

6. 

2+ 

a; 

X 

y 

a  _ 

y 

ac 
'  b 

M 
a 

■ 

166 


ALGEBRA. 


7.1  +  1  =  5 

ax     by 


ax 


2_ 
by 


8. 1 z=m-\-n 

nx     /my 

x      y 


9.M 

X 


—  m  I 

y 
y       ^ 


167.  If  three  or  more  simultaneous  equations  are  given, 
involving  three  or  more  unknown  numbers,  one  of  the 
unknowns  must  be  eliminated  between  two  or  Tuore  pairs 
of  the  equations ;  then  a  second  unknown  between  the 
pairs  that  can  be  formed  of  the  resulting  equations. 

Note.  The  pairs  chosen  to  eliminate  from  must  be  independent 
pairs,  so  that  each  of  the  given  equations  shall  be  used  in  the  process 
of  the  eliminations. 


Solve:  2a;-  32/  + 42=    4 

3a:  +  52/-7z  =  12 
6x—    y  —  8z=    5 

Eliminate  z  between  the  equations  (1)  and  (3). 
Multiply  (1)  by  2,       4cx-6y  +  Sz=    8 
(3)  is  5x-    y-Sz=   5 

Add, 


0) 
(2) 
(3) 

(4) 
(5) 


9x-7y  =13 

Eliminate  z  between  the  equations  (1)  and  (2). 
Multiply  (1)  by  7,  14x- 21^  +  28z  =  28 
Multiply  (2)  by  4,   12a;  +  20y  -  282  =  48 

Add,  26  a;-      y  =76  (6) 

"We  now  have  two  equations  (5)  and  (6)  involving  two  unknowns, 
X  and  y. 

Multiply  (6)  by  7,  182  a;  -  7  y  =  532  (7) 

(5)  is  9a;-7y=    13 

Subtract,  173  a;  =519 

.-.a;  =  3. 
Substitute  the  value  of  a;  in  (6),  78  —  y  =  76. 

•.y  =  2. 

Substitute  the  values  of  x  and  y  in  (1), 
6-6  +  4^  =  4. 

.-.3  =  1. 


SIMULTANEOUS   EQUATIONS.  167 

Exercise  68. 


Solve: 


1.    5x-{-Sy-6z  =  4:^         10.    3x-y-\-z=l7 
Sx-yi-2z  =  S     \  5xi'Sy-2z=10 


x-2y  +  2z  =  2    )  7x-{-4:y-^z  =  S 

2.  4:X-6y-{-2z  =  6    '\       11.  x-{-i/  +  z  =  b 

2x  +  S7/~z  =  20     i  3a;-5y+7z  =  75 

7a;_4y-f3z  =  35j  9a; -11^  +  10  =  0 

3.  a;  +  y  +  z  =  6               ]    12.  x  +  2y  +  ^z  =  6 
6x  +  4:y  +  Sz  =  22      i  2x  +  4:7/  +  2z  =  8 


15^;  +  lOy  +  62  =  53  J  Sx  +  2y-\-8z  =  10l  ) 

4.    4:X-Sy  +  z  =  9    -)  13.    x-Sy-2z  =  l 

9x  +  y-5z  =  16i  2a;-3y  +  52  =  -19 

a;-4y  +  3z  =  2   ]  5xi-2y-z=12 

a;  +  3y-2  =  7        I  4a;-32/  +  2z=ll  > 


4a:-5y  +  42  =  8J  x-2y~5z  =  -1 

6.  12a; +  5y-- 42  =  29-)  15.  a;  +  y  =  l 
13a;  —  2y +  52  =  58  I  y+z  =  9 
17a;  — y  — 2  =  15       J  a;+2  =  5 

7.  y-a;  +  z  =  — 5  ^  16.  2a;  — 3y  =  3 
2_y_a;  =  -25  [  3y-42  =  7 
a;-fy  +  z  =  35      J                   42-5a;  =  2 

8.  a; +  2/  + 2  =  30  ]  17.    3a;-4y  +  62  =  1 
8a;  +  42/  +  2z  =  50    >  2a;  +  2y-2  =  l 
27a;  +  9y  +  32  =  64]  7a; -63/ +  72  =  2 

9.  152/  =  242-10a;  +  41  -j  18.  7a;-32/  =  30 
15a;  =  12y-162  +  10  }-  9y-52=34 
18a;-(72-133  =  142/]            a;  +  y  +  2=33 


168 


19-   ^  +  1  +  1=6 


^+1+1- 

^+1+1=1^ 

20. 

1+2=5      1 

X     y 

§-4^-6 

y    z 

l-i  =  5 

2     a; 

1,11 

21. 

-  H =  a 

X     y      z 

1-M  =  A 

a;     2/      2 

1,1      1 

--] =  c 

y     z    X 

22. 

hz-\-cy-=a\ 

az-\-cx  =  b  If 

ay-{-bx=  c) 

23. 

--f+-=^ 

*     1 

X      by      z 

^+;^  +  -  =  10i 

4        1    ,  4      ifii 

24. 

2-§  +  4  =  2.9        1 

a;     y      2 

x     y      z 

2+10_8^149 

y      z     X             ) 

25. 

?+l_§=o1 

§-2-2=0 

2     y 

1+1-1=0 

a;      2      3         J 

26.  ax -{- hy -\- cz  =  a 
ax  —  by  —  cz=  b 
ax-[-  cy-\-bz=  c 


2x-y^^y^2z^x-y-z^^ 
3  4  6 


28. 


i=y- 


x-\-  z 


X  —  a  —  b 

a-\-b  -\-  c 


*  Subtract  from  the  sum  of  the  three  equations  each  equation  separately. 
f  Multiply  the  equations  by  a,  h,  and  c,  respectively,  and  from  the  sum 
of  the  results  subtract  the  double  of  each  equation  separately. 


CHAPTER  XII. 

PROBLEMS    INVOLVING   TVi^O    UNKNOWN 
NUMBERS. 

168.  It  is  often  necessary  in  the  solution  of  problems  to 
employ  two  or  more  letters  to  represent  the  numbers  to  be 
found.  In  all  cases  the  conditions  must  be  sufficient  to 
give  just  as  many  equations  as  there  are  unknown  numbers 
to  be  found. 

169.  If  there  are  more  equations  than  unknown  numbers, 
some  of  them  are  superfluous  or  inconsistent ;  if  there  are 
less  equations  than  unknown  numbers,  the  problem  is  inde- 
terminate. 

(1)  If  A  gives  B  $10,  B  will  have  three  times  as  much 
money  as  A.  If  B  gives  A  $  10,  A  will  have  twice  as  much 
money  as  B.     How  much  has  each  ? 

Let  X  =  number  of  dollars  A  has, 

and  y  =  number  of  dollars  B  has. 

Then,  after  A  gives  B  1 10, 

x  —  \0  =  the  number  of  dollars  A  has, 
2/  +  10  =  the  number  of  dollars  B  has. 
.-.  y  +  10  =  3(.T-10).  (1) 

If  B  gives  A  1 10, 

ic  +  10  =  the  number  of  dollars  A  has, 
y  —  10  =  the  number  of  dollars  B  has. 
.-.  a;  +  10  =  2(y-10).  (2) 

From  the  solution  of  equations  (1)  and  (2),  x  =  22,  and  y  =  26. 

Therefore  A  has  $22,  and  B  has  $26. 


170  ALGEBRA. 


Exercise  69. 


1.  The  sum  of  two  numbers  divided  by  2  gives  as  a  quo- 

tient 24,  and  the  difference  between  them  divided 
by  2  gives  as  a  quotient  17.  What  are  the  num- 
bers? 

2.  The  number  144  is  divided  into  three  numbers.   When 

the  first  is  divided  by  the  second,  the  quotient  is  3 
and  the  remainder  2 ;  and  when  the  third  is  divided 
by  the  sum  of  the  other  two  numbers,  the  quotient 
is  2  and  the  remainder  6.     Find  the  numbers. 

3.  Three  times  the  greater  of  two  numbers  exceeds  twice 

the  less  by  10 ;  and  twice  the  greater  together  with 
three  times  the  less  is  24.     Find  the  numbers. 

4.  If  the  smaller  of  two  ifitimbers  is  divided  by  the  greater, 

the  quotient  is  0.21  and  the  remainder  0.0057 ;  but 
if  the  greater  is  divided  by  the  smaller,  the  quo- 
tient is  4  and  the  remainder  0.742.  What  are  the 
numbers  ? 

5.  Seven  years  ago  the  age  of  a  father  was  four  times 

that  of  his  son ;  seven  years  hence  the  age  of  the 
father  will  be  double  that  of  the  son.  What  are 
their  ages  ? 

6.  The  sum  of  the  ages  of  a  father  and  son  is  half  what  it 

will  be  in  25  years ;  the  difference  between  their 
ages  is  one-third  of  what  the  sum  will  be  in  20  years. 
What  are  their  ages  ? 

7.  If  B  gives  A  f  25,  they  will  have  equal  sums  of  money ; 

but  if  A  gives  B  $  22,  B's  money  will  be  double  that 
of  A.     How  much  has  each  ? 


PROBLEMS.  171 

8.  A  farmer  sold  to  one  person' 30  bushels  of  wheat  and 

40  bushels  of  barley  for  $67.50;  to  another  person 
he  sold  50  bushels  of  wheat  and  30  bushels  of  barley 
for  $85.  What  was  the  price  of  the  wheat  and  of 
the  barley  per  bushel  ? 

9.  If  A  gives  B  $5,  he  will  then  have  $6  less  than  B; 

but  if  he  receives  $5  from  B,  three  times  his  money 
will  be  $20  more  than  four  times  B's.  How  much 
has  each  ? 

10.  The  cost  of  12  horses  and  14  cows  is  $1900  ;  the  cost 

of  5  horses  and  3  cows  is  $  650.  What  is  the  cost 
of  a  horse  and  a  cow  respectively  ? 

Note.    A  fraction  of  which  the  terms  are  unknown  may  be  rep- 
resented by  -. 

■y 

Ex.  A  certain  fraction  becomes  equal  to  |-  if  3  is  added  to 
its  numerator,  and  equal  to  -^  if  3  is  added  to  its 
denominator.     Determine  the  fraction. 

Let  -  =  the  required  fraction. 

y 

Then  ^-±^  =  I,  and  ^-  -  f . 

y  2/  +  3 

From  the  solution  of  these  equations  it  is  found  that 

a;  =  6;  *" 

2/  =  18. 

Therefore  the  fraction  ==  y^. 

11.  A  certain  fraction  becomes  equal  to  2  when  7  is  added 

to  its  numerator,  and  equal  to  1  when  1  is  subtracted 
from  its  denominator.     Determine  the  fraction. 

12.  A  certain  fraction  becomes  equal  to  \  when  7  is  added 

to  its  denominator,  and  equal  to  2  when  13  is  added 
to  its  numerator.     Determine  the  fraction. 


172  ALGEBRA. 

13.  A  certain  fraction  becomes  equal  to  -J  when  the  denom- 

inator is  increased  by  4,  and  equal  to  ff-  when  the 
numerator  is  diminished  by  15.  Determine  the 
fraction. 

14.  A  certain  fraction  becomes  equal  to  -|  if  7  is  added  to 

the  numerator,  and  equal  to  f  if  7  is  subtracted 
from  the  denominator.     Determine  the  fraction. 

15.  Find  two  fractions  with  numerators  2  and  5  respec- 

tively, the  sum  of  which  is  1-| ;  but  if  their  denomi- 
nators are  interchanged  the  sum  of  the  fractions  is  2. 

16.  A  fraction  which  is  equal  to  f  is  increased  to  ^  when 

a  certain  number  is  added  to  both  its  numerator  and 
denominator,  and  is  diminished  to  f  when  one  more 
than  the  same  number  is  subtracted  from  each. 
Determine  the  fraction. 

Note.  A  number  consisting  of  two  digits  which  are  unknown 
may  be  represented  by  10  a;  +  y,  in  which  x  and  y  represent  the  digits 
of  the  number.  Likewise,  a  number  consisting  of  three  digits  which 
are  unknown  may  be  represented  by  100  a;  +  lOy  +  z,  in  which  x,  y, 
and  z  represent  the  digits  of  the  number. 

For  example,  consider  any  number  expressed  by  three  digits,  as 
364.  The  expressiga  364  mea^iis  300  +  QO  +  4 ;  or,  100  times  3  +  10 
times  6  +  4. 

Ex.  The  sum  of  the  two  digits  of  a  number  is  8,  and  if  36 
be  added  to  the  number,  the  digits  will  be  inter- 
changed.    What  is  the  number? 

Let  X  =  the  digit  in  the  tens'  place, 

and  y  =  the  digit  in  the  units'  place. 

Then  10a;  +  y  =  the  number. 

By  the  conditions,  x  +  y  =  8,  (1) 

and  10a;  +  2/  +  36=10y +  a;.  (2) 

From  (2),  9a; -91/  = -36. 

Divide  by  9,  a?  -  2/  =  -  4.  (^) 


PROBLEMS.  173 

Add  (1)  and  (3),  2a;  =4. 

.-.  x  =  2. 
Subtract  (3)  from  (1),  2^  =  12. 

.-.  y  =  6. 
Hence  the  number  is  26. 

17.  The  sum  of  the  two  digits  of  a  number  is  10,  and  if  54 

be  added  to  the  number  the  digits  will  be  inter- 
changed.    What  is  the  number  ? 

18.  The  sum  of  the  two  digits  of  a  number  is  6,  and  if  the 

number  is  divided  by  the  sum  of  the  digits  the  quo- 
tient is  4.     What  is  the  number  ? 

19.  A  certain  number  is  expressed  by  two  digits,  of  which 

the  tens'  digit  is  the  greater.  If  the  number  is 
divided  by  the  sum  of  its  digits  the  quotient  is  7 ; 
if  the  digits  are  interchanged,  and  the  resulting 
number  diminished  by  12  is  divided  by  the  differ- 
ence between  the  two  digits,  the  quotient  is  9. 
What  is  the  number? 

20.  If  a  certain  number  is  divided  by  the  sum  of  its  two 

digits  the  quotient  is  6  and  the  remainder  3  ;  if  the 
digits  are  interchanged,  and  the  resulting  number  is 
divided  by  the  sum  of  the  digits,  the  quotient  is  4 
and  the  remainder  9.     What  is  the  number  ? 

21.  If  a  certain  number  is  divided  by  the  sum  of  its  two 

digits  diminished  by  2,  the  quotient  is  5  and  the 
remainder  1 ;  if  the  digits  are  interchanged,  and  the 
resulting  number  is  divided  by  the  sum  of  the  digits 
increased  by  2,  the  quotient  is  5  and  the  remainder 
8.     Find  the  number. 

22.  The  first  of  the  two  digits  of  a  number  is,  when  doubled, 

3  more  than  the  second,  and  the  number  itself  is  less 
by  6  than  5  times  the  sum  of  the  digits.  What 
is  the  number  ? 


174  ALGEBRA. 

23.  A  number  is  expressed  by  three  digits,  of  which  the 

first  and  last  are*  alike.  By  interchanging  the  digits 
in  the  units'  and  tens'  places  the  number  is  increased 
by  54  ;  but  if  the  digits  in  the  tens'  and  hundreds' 
places  are  interchanged,  9  must  be  added  to  four 
,  times  the  resulting  number  to  make  it  equal  to  the 
I  original  number.     What  is'  the  number  ? 

24.  A  number  is  expressed  by  three  digits.  The  sum  of 
the  digits  is  21 ;  the  sum  of  the  first  and  second 
exceeds  the  third  by  3 ;  and  if  198  be  added  to  the 

number,  the  digits  in  the  units'  and  hundreds'  places 
will  be  interchanged.     Find  the  number. 

25.  A  number  is  expressed  by  three  digits.     The  sum  of 

the  digits  is  9 ;  the  number  is  equal  to  forty-two 
times  the  sum  of  the  first  and  second  digits ;  and 
the  third  digit  is  twicQ  the  sum  of  the  other  two. 
Find  the  number. 

26.  A  certain  number,  expressed  by  three  digits,  is  equal 

to  forty-eight  times  the  sum  of  its  digits.  If  198  be 
subtracted  from  the  number,  the  digits  in  the  units' 
and  hundreds'  places  will  be  interchanged  ;  and  the 
sum  of  the  extreme  digits  is  equal  to  twice  the  mid- 
dle digit.     Find  the  number. 

Note.  If  a  boat  moves  at  the  rate  of  x  miles  an  hour  in  still 
"water,  and  if  it  is  on  a  stream  that  runs  at  the  rate  of  y  miles  an 
hour,  then 

X  +  y  represents  its  rate  down  the  stream, 
x  —  y  represents  its  rate  up  the  stream. 

27.  A  waterman  rows  30  miles  and  back  in  12  hours.    He 

finds  that  he  can  row  5  miles  with  the  stream  in  the 
same  time  as  3  against  it.  Find  the  time  it  takes 
him  to  row  up  and  down  respectively. 


PROBLEMS.  175 

28.  A  crew  which  can  pull  at  the  rate  of  12  miles  an  hour 

down  the  stream,  finds  that  it  takes  twice  as  long  to 
come  up  the  river  as  to  go  down.  At  what  rate 
does  the  stream  flow  ? 

29.  A  man  sculls  down  a  stream,  which  runs  at  the  rate 

of  4  miles  an  hour,  for  a  certain  distance  in  1  hour 
and  40  minutes.  In  returning  it  takes  him  4  hours 
and  15  minutes  to  arrive  at  a  point  3  miles  short  of 
his  starting-place.  Find  the  distance  he  pulled 
down  the  stream  and  the  rate  of  his  pulling. 

30.  A  person  rows  down  a  stream  a  distance  of  20  miles 

and  back  again  in  10  hours.  He  finds  he  can  row 
2  miles  against  the  stream  in  the  same  timef  he  can 
row  3  miles  with  it.  Find  the  time  of  his  rowing 
down  and  of  his  rowing  up  the  stream  ;  and  also  the 
rate  of  the  stream. 

Note.  When  commodities  are  mixed,  the  quantity  of  the  mixture 
is  equal  to  the  quantity  of  the  ingredients ;  the  cost  of  the  mixture 
is  equal  to  the  cost  of  the  ingredients. 

Ex.  A  wine-merchant  has  two  kinds  of  wine  which  cost 
72  cents  and  40  cents  a  quart  respectively.  How 
much  of  each  must  he  take  to  make  a  mixture  of  50 
quarts  worth  60  cents  a  quart  ? 

Let  X  =  required  number  of  quarts  worth  .72  cents  a 

quart, 
and  y  =  required  number  of  quarts  worth  40  cents  a 

quart. 
Then,  72aj  =  cost  in  cents  of  the  first  kind, 

4O3/  =  cost  in  cents  of  the  second  kind  of  wine, 
and  3000  =  cost  in  cents  of  the  mixture. 

.-.  x  +  y  =  50, 
72  a: +  40?/ =  3000. 
From  which  equations  the  values  of  x  and  y  may  be  found. 


176  ALGEBRA. 

31.  A  grocer  mixed  tea  that  cost  him  42  cents  a  pound 

with  tea  that  cost  him  54  cents  a  pound.  He  had  30 
pounds  of  the  mixture,  and  by  selling  it  at  the  rate 
of  60  cents  a  pound,  he  gained  as  much  as  10  pounds 
of  the  cheaper  toa  cost  him.  How  many  pounds  of 
each  did  he  put  into  the  mixture  ? 

32.  A  grocer  mixes  tea  that  cost  him  90  cents  a  pound 

with  tea  that  cost  him  28  cents  a  pound.  The  cost 
of  the  mixture  is  $61.20.  He  sells  the  mixture  at 
50  cents  a  pound,  and  gains  $3.80.  How  many 
pounds  of  each  did  he  put  into  the  mixture  ? 

33.  A  farmer  has  28  bushels  of  barley  worth  84  cents  a 

bushel.  With  his  barley  he  wishes  to  mix  rye  worth 
$1.08  a  bushel,  and  wheat  worth  $1.44  a  bushel,  so 
that  the  mixture  may  be  100  bushels,  and  be  worth 
$1.20  a  bushel.  How  many  bushels  of  rye  and  of 
wheat  must  he  take  ? 

Ex.  A  and  B  together  can  do  a  piece  of  work  in  48  days ; 
A  and  0  together  can  do  it  in  30  days ;  B  and  C 
together  can  do  it  in  26f  days.  How  long  will  it 
take  each  to  do  the  work  ? 

Let  X  =  the  number  of  days  it  will  take  A  alone  to  do  the  work, 

y  =  the  number  of  days  it  will  take  B  alone  to  do  the  work, 

and      z  =  the  number  of  days  it  will  take  C  alone  to  do  the  work. 

Then,  -,  -,  -,  respectively,  will  denote  the  part  each  can  do 
^        in  a  day, 

and  -  +  -  will  denote  the  part  A  and  B  together  can  do  in  a  day  ; 
X     y 

but  —  will  denote  the  part  A  and  B  together  can  do  in  a  day. 
Therefore,  1  + 1  =  J_.  (1) 


PROBLEMS.  177 

Likewise,  -  +  -  =  ^'  (2) 

and  -  +  -  =  ^z  (^) 

y     z     26§ 

The  solution  of  these  equations  gives  120,  80,  and  40  for  the 
values  of  x,  y,  and  z,  respectively. 

34.  A  and  B  together  earn   $40   in  6  days;    A  and  C 

together  earn  $54  in  9  days  ;  B  and  C  together 
earn  $80  in  15  days.     What  does  each  earn  a  day? 

35.  A  cistern  has  three  pipes,  A,  B,  and  C.     A  and  B  will 

fill  it  in  1  hour  and  10  minutes ;  A  and  C  in  1  hour . 
and  24  minutes ;  B  and  C  in  2  hours  and  20  minutes. 
How  long  will  it  take  each  to  fill  it  ? 

36.  A  warehouse  will  hold  24  boxes  and  20  bales ;  6  boxes 

and  14  bales  will  fill  half  of  it.  How  many  of  each 
alone  will  it  hold  ? 

37 .  Two  workmen  together  complete  some  work  in  20  days ; 

but  if  the  first  had  worked  twice  as  fast,  and  the 
second  half  as  fast,  they  would  have  finished  it  in 
15  days.  How  long  would  it  take  each  alone  to  do 
the  work  ? 

38.  A  purse  holds  19  crowns  and  6  guineas ;  4  crowns  and 

5  guineas  fill  JJ  of  it.  How  many  of  each  alone  will 
it  hold  ? 

39.  A  piece  of  work  can  be  completed  by  A,  B,  and  C 

together  in  10  days;  by  A  and  B  together  in  12 
days;  by  B  and  C,  if  B  work  15  days  and  0  30 
days.  How  long  w^ill  it  take  each  alone  to  do  the 
work  ? 

40.  A  cistern  has  three  pipes,  A,  B,  and  C.     A  and  B  will 

fill  it  in  a  minutes ;  A  and  C  in  6  minutes ;  B  and 
C  in  c  minutes.  How  long  will  it  take  each  alone 
to  fill  it? 


178  ALGEBRA. 

Note.  In  considering  the  rate  oj  increase  or  decrease  in  quantities, 
it  is  usual  to  take  100  as  a  common  standard  of  reference,  so  that  the 
increase  or  decrease  is  calculated  for  every  100,  and  therefore  called 
per  cent. 

The  representative  of  the  number  resulting  after  an  increase  has 
taken  place  is  100  +  increase  per  cent;  and  after  a  decrease,  100  — 
decrease  per  cent. 

Interest  depends  upon  the  time  for  which  the  money  is  lent,  as 
well  as  upon  the  rate  per  cent  charged ;  the  rate  per  cent  charged 
being  the  rate  per  cent  on  the  principal  for  one  year.     Hence, 

Simple  interest  =  ^^ncipal  X  Rate  per  cent  x  Time^ 
^  100 

where  Time  means  number  of  years  or  fraction  of  a  year. 
Amount  =  Principal  +  Interest. 
In  questions  relating  to  stocks,  100  is  taken  as  the  representative 
of  the  stock,  the  price  represents  its  market  value,  and  the  per  cent 
represents  the  interest  which  the  stock  bears.  Thus,  if  six  per  cent 
stocks  are  quoted  at  108,  the  meaning  is,  that  the  price  of  1 100  of 
the  stock  is  $108,  and  that  the  interest  derived  from  $100  of  the  stock 
will  be  jf ^  of  $  100 ;  that  is,  1 6  a  year.  The  rate  of  interest  on  the 
money  invested  will  be  |f  f  of  6  per  cent. 

41.  A  man  has  $10,000  invested.     For  a  part  of  this  sum 

he  receives  5  per  cent  interest,  and  for  the  rest  4 
per  cent ;  the  income  from  his  5  per  cent  investment 
is  $50  more  than  from  his  4  per  cent.  How  much 
has  he  in  each  investment  ? 

42.  A  sum  of  money,  at  simple  interest,  amounted  in  6 

years  to  $26,000,  and  in  10  years  to  $30,000.  Find 
the  sum  and  the  rate  of  interest. 

43.  A  sum  of  money,  at  simple  interest,  amounted  in  10 

months  to  $26,250,  and  in  18  months  to  $27,250. 
Find  the  sum  and  the  rate  of  interest. 

44.  A  sum  of  money,  at  simple  interest,  amounted  in  m 

years  to  a  dollars,  and  in  n  years  to  h  dollars.  Find 
the  sum  and  the  rate  of  interest. 


PROBLEMS.  179 

45.  A  sum  of  money,  at  simple  interest,  amounted  in  a 

months  to  c  dollars,  and  in  b  months  to  d  dollars. 
Find  the  sum  and  the  rate  of  interest. 

46.  A  person  has  a  certain  capital  invested  at  a  certain 

rate  per  cent.  Another  person  has  $1000  more 
capital,  and  his  capital  invested  at  one  per  cent 
better  than  the  first,  and  receives  an  income  $80 
greater.  A  third  person  has  $1500  more  capital, 
and  his  capital  invested  at  two  per  cent  better  than 
the  first,  and  receives  an  income  $150  greater. 
Find  the  capital  of  each,  and  the  rate  at  which  it  is 
invested. 

47.  A  person  has  $12,750  to  invest.     He  can  buy  three 

per  cent  bonds  at  81,  and  five  per  cents  at  120. 
Find  the  amount  of  money  he  must  invest  in  each  in 
order  to  have  the  same  income  from  each  invest- 
ment. 

Hint.  If  x  and  y  represent  the  number  of  dollars  invested  in  the 
three  and  five  per  cents  respectively,  then  —-  and  — ^  are  the 
respective  incomes  from  the  three  and  five  per  cents. 

48.  A  and  B  each  invested  $1500  in  bonds;  A  in  three 

per  cents  and  B  in  four  per  cents.  The  bonds  were 
bought  at  such  prices  that  B  received  $  5  interest 
more  than  A.  After  both  classes  of  bonds  rose  10 
points,  they  sold  out,  and  A  received  $50  more  than 
B.     What  price  was  paid  for  each  class  of  bonds  ? 

Hint.     If  x  and  y  represent  the  cost  of  $  1  in  three  per  cents  and  $  1 

in  four  per  cents,  respectively,  then  — —  and  — — —  are  the  face  values 

3       1500 
0^  the  three  and  four  per  cents,  respectively  ;  and  — —  X and 

4        1500  ^ 

rJL^  X are  the  respective  incomes. 

100        V 


180  ALGEBRA. 

49.  A  person  invests  $10,000  in  three  per  cent  bonds, 
$16,500  in  three  and  one-half  per  cents,  and  has  an 
income  from  both  investments  of  $1056.25.  If  his 
investments  had  been  $2750  more  in  the  three  per 
cents,  and  less  in  the  three  and  one-half  per  cents, 
his  income  would  have  been  62J  cents  greater. 
What  price  was  paid  for  each  class  of  bonds  ? 

Hint.     Let  x  and  y  represent  the  cost  of  $  1  in  three  and  three 

and   one-half  per  cent  bonds  respectively ;  then  x and 

-^  X are  the  respective  incomes. 


100 


y 


50.  The  sum  of  $2500  was  divided  into  two  unequal  parts 

and  invested,  the  smaller  part  at  two  per  cent  more 
than  the  larger.  The  rate  of  interest  on  the  larger 
sum  was  afterwards  increased  by  1,  and  that  on  the 
smaller  sum  diminished  by  1 ;  and  thus  the  interest 
of  the  whole  was  increased  by  one-fourth  of  its  value. 
If  the  interest  of  the  larger  sum  had  been  so  in- 
creased, and  no  change  been  made  in  the  interest  of 
the  smaller  sum,  the  interest  of  the  whole  would 
have  been  increased  one-third  of  its  value.  Find 
the  sums  invested,  and  the  rate  per  cent  of  each. 

Note.  If  a;  represents  the  number  of  linear  units  in  the  length, 
and  y  in  the  width,  of  a  rectangle,  xy  represents  the  number  of  its 
units  of  surface ;  the  surface  unit  having  the  same  name  as  the  linear 
unit  of  its  sides. 

51.  If  the  sides  of  a  rectangular  field  were  each  increased 

by  2  yards,  the  area  would  be  increased  by  220 
square  yards  ;  if  the  length  were  increased  and  the 
breadth  were  diminished  each  by  5  yards,  the  area 
would  be  diminished  by  185  square  yards.  What 
is  its  area  ? 


PROBLEMS.  181 

52.  If  a  given  rectangular  floor  had  been  3  feet  longer  and 

2  feet  broader  it  would  have  contained  64  square 
feet  more ;  but  if  it  had  been  2  feet  longer  and  3 
feet  broader  it  would  have  contained  68  square  feet 
more.     Find  the  length  and  breadth  of  the  floor. 

53.  In  a  certain  rectangular  garden  there  is  a  strawberry- 

bed  whose  sides  are  one-third  of  the  lengths  of  the 
corresponding  sides  of  the  garden.  The  perimeter  of 
the  garden  exceeds  that  of  the  bed  by  200  yards ; 
and  if  the  greater  side  of  the  garden  be  increased  by 
3,  and  the  other  by  5  yards,  the  garden  will  be  en- 
larged by  645  square  yards.  Find  the  length  and 
breadth  of  the  garden. 

Note.     Care  must  be  taken  to  express  the  conditions  of  a  problem 
in  the  same  principal  unit. 

Ex.  In  a  mile  race  A  gives  B  a  start  of  20  yards  and  beats 
him  by  30  seconds.  At  the  second  trial  A  gives  B  a 
start  of  32  seconds  and  beats  him  by  9y\  yards. 
Find  the  rate  per  hour  at  which  each  runs. 

Let  X  =  number  of  yards  A  runs  a  second, 
and      y  =  number  of  yards  B  runs  a  second. 
Since  there  are  1760  yards  in  a  mile, 

=  number  of  seconds  it  takes  A  to  run  a 

^  mile. 

and  ~ — 3i  =  number  of  seconds  B  was  running  in  the 

y  y  first  and  second  trials,  respectively. 

Hence,  mo -lip  =30, 

and        IJiOA- 1^=32. 

y  ^ 

The  solution  of  these  equations  gives  x  =  5|-f  and  y  =•  5/j-. 

513  1 

That  is,  A  runs  — ^,  or  — ,  of  a  mile  in  one  second  ; 

1760        300 

and  in  one  hour,  or  3600  seconds,  runs  12  miles. 
Likewise,  B  runs  lOy^j  miles  in  one  hour. 


182  ALGEBRA. 

54.  In  a  mile  race  A  gives  B  a  start  of  100  yards  and  beats 

him  by  15  seconds.  In  the  second  trial  A  gives  B  a 
start  of  45  seconds  and  is  beaten  by  22  yards.  Find 
the  rate  of  each  in  miles  per  hour. 

55.  In  a  mile  race  A  gives  B  a  start  of  44  yards  and  beats 

him  by  51  seconds.  In  the  second  trial  A  gives  B  a 
start  of  1  minute  and  15  seconds  and  is  beaten  by  88 
yards.     Find  the  rate  of  each  in  miles  per  hour. 

56.  The  time  which  an  express  train  takes  to  go  120  miles 

is  Y^^  of  the  time  taken  by  an  accommodation  train. 
The  slower  train  loses  as  much  time  in  stopping  at 
different  stations  as  it  would  take  to  travel  20  miles 
without  stopping ;  the  express  train  loses  only  half 
as  much  time  by  stopping  as  the  accommodation 
train,  and  travels  15  miles  an  hour  faster.  Find  the 
rate  of  each  train  in  miles  per  hour. 

Hint.    If  x  and  y  represent  the  rates  of  the  slower  and  faster  trains 

120         I'^O 
respectively,  then and  -JL-  represent  the  number  of  hours  it  takes 

•   ^  2/  20  10 

the  respective  trains  to  run  120  miles ;  —  and  —  represent  the  num- 

X  X 

ber  of  hours  the  respective  trains  lose  by  stopping. 

57.  A  train  moves  from  P  towards  Q,  and  an  hour  later  a 

second  train  starts  from  Q  and  moves  towards  P 
at  a  rate  of  10  miles  an  hour  more  than  the  first 
train ;  the  trains  meet  half-way  between  P  and  Q. 
If  the  train  from  P  had  started  an  hour  after  the 
train  from  Q,  its  rate  must  have  been  increased  by  28 
miles  in  order  that  the  trains  should  meet  at  the 
half-way  point.     Find  the  distance  from  P  to  Q. 

Hint.  If  x  denotes  the  number  of  hours  it  takes  the  first  train  to 
go  half  the  distance,  and  y  denotes  the  rate  of  the  first  train,  then 
X  —  1  denotes  the  number  of  hours  it  takes  the  second  train  to  go 
half  the  distance,  and  y  +  \0  denotes  the  rate  of  the  second  train. 
Hence,  xy  and  {x  —V){y-\- 10)  are  each  equal  to  half  the  distance. 


PROBLEMS.  183 

58.  A  passenger  train,  after  travelling  an  hour,  meets  with 

an  accident  which  detains  it  one-half  an  hour  ;  after 

which  it  proceeds  at  four-fifths  of  its  usual  rate,  and 

arrives  an  hour  and  a  quarter  late.     If  the  accident 

had  happened  30  miles  farther  on,  the  train  would 

have  been  only  an  hour  late.     Determine  the  usual 

rate  of  the  train. 

Hint.     If  x  represents  the  number  of  miles  the  train  usually  goes 

an  hour,  and  y  the  whole  distance  in  miles,  3/  —  re  is  the  distance  the 

train  has  to  go  after  the  accident,  and  ^  ~  ■  is  the  number  of  hours 

usually  required,  and  ^-"^  is  the  number  of  hours  required  after  the 

accident.     Hence  ^-^^^ —  ^  ~     is  the  loss  in  running  time, 
^x  X 

Also,  since  the  detention  is  ^  hour,  and  the  train  is  IJ  hours  late, 

the  loss  in  running  time  is  |  of  an  hour. 

Therefore,  ^zJE  -  ^-=l:£  =  ^. 

^X  X  4: 

If  the  accident  had  happened  30  miles  farther  on,  the  distance  to 
be  run  would  have  been  y  —  {x  +  30),  and  we  should  have  had 
y-(x  +  30)     y  -  (x  +  30)  _  1 
|x  X  2 

59.  A  passenger  train  after  travelling  an  hour  is  detained 

15  minutes ;  after  which  it  proceeds  at  three-fourths 
of  its  former  rate,  and  arrives  24  minutes  late.  If 
the  detention  had  taken  place  5  miles  farther  on,  the 
train  would  have  been  only  21  minutes  late.  Deter- 
mine the  usual  rate  of  the  train. 

60.  A  man  bought  10  oxen,  120  sheep,  and  46  lambs. 

The  cost  of  3  sheep  was  equal  to  that  of  5  lambs ; 
an  ox,  a  sheep,  and  a  lamb  together  cost  a  number 
of  dollars  less  by  57  than  the  whole  number  of 
animals  bought;  and  the  whole  sum  spent  was 
$2341.50.  Find  the  price  of  an  ox,  a  sheep,  and  a 
lamb,  respectively. 


184  ALGEBRA. 

61.  A  farmer  sold  100  head  of  stock,  consisting  of  horses, 

oxen,  and  sheep,  so  that  the  whole  realized  $11.76  a 
head ;  while  a  horse,  an  ox,  and  a  sheep  were  sold 
for  1 110,  $62.50,  and  $7.50,  respectively.  Had  he 
sold  one-fourth  of  the  number  of  oxen  that  he  did, 
and  25  more  sheep,  he  would  have  received  the  same 
sum.  Find  the  number  of  horses,  oxen,  and  sheep, 
respectively,  which  were  sold. 

62.  A,  B,  and  0  together  subscribed  $100.     If  A's  sub- 

scription had  been  one-tenth  less,  and  B's  one-tenth 
more,  O's  must  have  been  increased  by  $2  to  make 
up  the  sum ;  but  if  A's  had  been  one-eighth  more, 
and  B's  one-eighth  less,  C's  subscription  would  have 
been  $17.50.     What  did  each  subscribe? 

63.  A  gives  to  B  and  0  as  much  as  each  of  them  has ;  B 

gives  to  A  and  0  as  much  as  each  of  them  then  has ; 
and  0  gives  to  A  and  B  as  much  as  each  of  them 
then  has.  In  the  end  each  of  them  has  $6.  How 
much  had  each  at  first  ? 

64.  A  pays  to  B  and  0  as  much  as  each  of  them  has;  B 

pays  to  A  and  0  one-half  as  much  as  each  of  them 
then  has ;  and  0  pays  to  A  and  B  one-third  of  what 
each  of  them  then  has.  In  the  end  A  finds  that  he 
has  $1.50,  B  $4.16|,  C  $0,581  How  much  had 
each  at  first  ? 

170.  Discussion  of  Problems.  The  discussion  of  a  problem 
consists  in  making  various  suppositions  as  to  the  relative 
values  of  the  given  numbers,  and  explaining  the  results. 
We  will  illustrate  by  an  example : 

Two  couriers,  A  and  B,  were  travelling  along  the  same 
road,  and  in  the  same  direction,  from  0  towards  D.  A 
travels  at  the  rate  of  m  miles  an  hour,  and  B  at  the  rate 


PROBLEMS.  185 

of  n  miles  an  hour.  At  12  o'clock  B  was  d  miles  in 
advance  of  A.     When  will  the  couriers  be  together  ? 

Suppose  they  will  be  together  x  hours  after  12,  Then  A  has  trav- 
elled mx  miles,  and  B  has  travelled  nx  miles,  and  as  A  has  travelled 
d  miles  more  than  B, 

.•.  mx  =  nx  -\-  d, 
or  mx  —  nx  =  d. 

m  —  n 

Discussion  of  the  Problem.     1.  If  m  is  greater  than  n,  the  value 

of  X,  namely, ,  is  positive,  and  it  is  evident  that  A  will  over- 

m  —  n 
take  B  after  12  o'clock. 

2,  If  m  is  less  than  n,  then will  be  negative.     In  this  case 

m,  —  n 

B  travels  faster  than  A,  and  as  he  is  d  miles  ahead  of  A  at  12  o'clock 
it  is  evident  that  A  cannot  overtake  B  after  12  o'clock,  but  that  B 
passed  A  before  12  o'clock.  The  supposition,  therefore,  that  the 
couriers  were  together  after  12  o'clock  was  incorrect,  and  the  negative 
value  of  X  points  to  an  error  in  the  supposition. 

3,  If  m  equals  n,  then  the  value  of  x,  that  is, ,  assumes  the 

7  7n  —  n 

form  -.     Now  if  the  couriers  were  d  miles  apart  at  12  o'clock,  and  if 

they  had  been  travelling  at  the  same  rates,  and  continue  to  travel  at 
the  same  rates,  it  is  obvious  that  they  never  had  been  together,  and 
that  they  never  will  be  together,  so  that  the  symbol  -  may  be  regarded 
as  the  symbol  of  impossibility. 

4,  If  7n  equals  n  and  d  is  0,  then  becomes  -.     Now  if  the 

m  —  n  0 

couriers  were  together  at  12  o'clock,  and  if  they  had  been  travelling 
at  the  same  rates,  and  continue  to  travel  at  the  same  rates,  it  is 
obvious  that  they  had  been  together  all  the  time,  and  that  they  will 
continue  to  be  together  all  the  time,  so  that  there  is  an  indefinite 
number  of  solutions.  Therefore,  the  symbol  -  may  be  regarded  as 
the  symbol  of  indetermination. 


CHAPTER  XIII. 

SIMPLE    INDETERMINATE   EQUATIONS. 

171.  If  a  single  equation  involving  two  unknown 
numbers  is  given,  and  no  other  condition  is  imposed,  the 
number  of  solutions  of  the  equation  is  unlimited  ;  for  if 
one  of  the  unknown  numbers  be  assumed  to  have  any 
value,  a  corresponding  value  of  the  other  may  be  found. 

Such  an  equation  is  called  an  indeterminate  equation. 

Although  the  number  of  solutions  of  an  indeterminate 
equation  is  unlimited,  the  values  of  the  unknown  numbers 
are  confined  to  a  particular  range ;  this  range  may  be  fur- 
ther limited  by  some  restriction,  as  for  example  by  requir 
ing  that  the  unknown  numbers  shall  be  positive  integers. 

172.  Every  indeterminate  equation  of  the  first  degree, 
in  which  x  and  y  are  the  unknown  numbers,  may  be  made 
to  assume  the  form 

ax±b7/  =  ±:c, 

where  a,  b,  and  c  are  positive  integers  and  have  no  common 
factor. 

Note.  The  sign  ±  is  read  plus  or  minus,  and  the  sign  +  is  read 
minus  or  plus. 

173.  Examples. 

(1)   Solve  3a;  -f  4y  —  22,  in  positive  integers. 
Transpose,  3  a;  =  22  —  4  3/. 

the  quotient  being  written  as  a  mixed  expression. 


SIMPLE    INDETERMINATE   EQUATIONS.  187 

Since  the  values  of  x  and  y  are  to  be  integral,  x  -{■  y  —*1  will  be 

integral,  and  hence      ~  ^  will  be  integral,  though  written  in  the 
form  of  a  fraction. 

Let  ~y  =  m,  an  integer. 

Then  1  —  y  =  3  m. 

.'.  y  =  1  —  3m. 
Substitute  this  value  of  y  in  the  original  equation, 
3a; +  4 -12m  =  22. 

.-.  x  =  6  +  4m. 

The  equation  y  =  l  —  3m  shows  that  m  may  be  0,  or  have  any- 
negative  integral  value,  but  cannot  have  a  positive  integral  value. 

The  equation  a?  =  6  +  4  m  further  shows  that  m  may  be  0,  but  can- 
not have  a  negative  integral  value  greater  than  1. 

.'.  m  may  be  0  or  —  1 ; 
and  then  ic  =  6 )  x  =  2 

y  =  l}  2/  =  4 

(2)   Solve  5x~  14y  =  11,  in  positive  integers. 
Transpose,  5  a;  =  11  +  14  y, 

x=2  +  2y  +  l±^-  (!) 

o 

,..^_2y-2  =  l±i^. 
^  5 

Since  x  and  y  are  to  be  integral,  x  —  2y  —  2  will  be  integral,  and 

hence  —^ — ^  will  be  integral. 
5 

Let  — i — ^  =  m,  an  integer. 

5 

mi  5  m  — 1 

Then  y  =  — - — . 

or  V  y  =  m  +  — - —  (2) 

Now    ^  ~-  r-iust  be  integral. 


188  ALGEBRA. 

-1 

Let  ■  =  n,  an  integer. 

Then  m  =  4n  +  l. 

Substituting  in  (2),  y  =  5n  +  1. 

Substituting  in  (1),  a;  =  14n  +  5. 

Obviously  x  and  y  will  both  be  positive  integers  if  n  have  any 
positive  integral  value. 

Hence,  x  =  5,     19,     33,     47,     

y  =  l,      6,     11,     16,     

It  will  be  seen  from  (1)  and  (2)  that  when  only  positive  integers 
are  required,  the  number  of  solutions  will  be  limited  or  unlimited 
according  as  the  sign  connecting  x  and  y  is  positive  or  negative. 

(3)  Find  the  least  number  that  when  divided  by  14  and 
5  will  give  remainders  1  and  3  respectively. 

If  iV  represent  the  number,  then 

X,  and 


14  5         " 

.-.  iV=  14a;  +  1,  and  N==  5y  +  3. 

.•.Ux  +  l  =  5y  +  3. 

5y  =  Ux-2, 

by  =  15a;  —  2  —  a;. 

o         2  +  a; 

.-.  y  =?>x — 

^  5 

T   i.  2  +  a; 

Let  =  m,  an  integer 

5 

.-.  a;  =  5  m  — 2. 

y  =  \{l4iX  —  2),  from  the  original 
.•.  2/  =  14  m  —  6. 
Ifm  =  l,  a;  =3,  and  2/ =  8. 

.-.  i\^=14a;  +  l  =  5y  +  3  =  43. 

(4)  Solve  5 a; -|- 6y  =  30,  so  that  x  may  be  a  multiple  of  y, 
and  both  x  and  y  positive. 

Let  X  =  my. 

Then  (5  m +  6)  3/ =  30. 


SIMPLE    INDETERMINATE    EQUATIONS.  189 

30 


y  = 


5m  +  6 
30  m 


5m  +  6 
Ifm  =  2,  a;  =  3f,  t/  =  l|. 

Ifw  =  3,  x  =  ^,  2/  =  lf- 

(5)  Solve  \^x  -\-  22y  =  71,  in  positive  integers. 

^         14 
If  we  jnultiply  the  fraction  by  7  and  reduce,  the  result  is 

a  form  which  shows  that  there  can  be  no  integral  solution. 

There  can  be  no  integral  solution  of  ax±hy  =  ±Gii  a  and  h  have 
a  common  factor  not  common  also  to  c ;  for  if  c?  be  a  factor  of  a  and 
also  of  h,  but  not  of  c,  the  equation  may  be  written 

mdx  ±  ndy  =  ±  c,  or  nx  ±ny  =  ±-; 

which  is  impossible,  since  -  is  a  fraction,  and  mx  ±  ny  is  an  integer, 
if  X  and  y  are  integers. 

Exercise  70. 
Solve  in  positive  integers  : 

1.  2a: +112/ =  49.  5.    3a;  +  8y  =  61. 

2.  7a;  +  3y  =  40.  6.   8:r  +  5y  =  97. 

3.  5:r  +  7y  =  53.  7.    lQx  +  ly  =  nO. 

4.  a;+10y  =  29.  8.    7a;+10y  =  206. 
Solve  in  least  positive  integers : 

9.    I2x-1y  =  l.  12.   23a;  — 9y  =  929. 

10.  5a;  — 17y  =  23.  13.    23a;-33y  =  43. 

11.  22>y~l2,x  =  Z.  14.    555a;— 22y  =  73. 


190  ALGEBRA. 

15.  How  many  fractions  are  there  with  denominators  12 

and  18  whose  sum  is  ||  ? 

16.  What  is  the  least  number  which,  when  divided  by  3 

and  5,  leaves  remainders  2  and  3  respectively  ? 

17.  A  person  counting  a  basket  of  eggs,  which  he  knows 

are  between  50  and  60,  finds  that  when  he  counts 
them  3  at  a  time  there  are  2  over;  but  when  he 
counts  them  5  at  a  time  there  are  4  over.  How 
many  are  there  in  all  ? 

18.  A  person  bought  40  animals,  consisting  of  pigs,  geese, 

and  chickens,  for  $40.  The  pigs  cost  $5  apiece, 
the  geese  $1,  and  the  chickens  25  cents  each.  Find 
the  number  he  bought  of  each. 

19.  Find  the  least  multiple  of  7  which,  when  divided  by 

2,  3,  4,  5,  6,  leaves  in  each  case  1  for  a  remainder. 

20.  In  how  many  ways  may  100  be  divided  into  two  parts, 

one  of  which  shall  be  a  multiple  of  7  and  the  other 
of9? 

21.  Solve  18a;  —  53/  =  70  so  that  y  may  be  a  multiple  of 

X,  and  both  positive. 

22.  Solve  8:r  +  12y  =  23  so  that  x  and  y  may  be  positive, 

and  their  sum  an  integer. 

23.  Divide  70  into  three  parts  which  shall  give  integral 

quotients  when  divided  by  6,  7,  8,  respectively,  and 
the  sum  of  the  quotients  shall  be  10. 

24.  Divide  200  into  three  parts  which  shall  give  integral 

quotients  when  divided  by  5,  7, 11,  respectively,  and 
the  sum  of  the  quotients  shall  be  20. 

25.  A  number  consisting  of  three  digits,  of  which  the  mid- 

dle one  is  4,  has  the  digits  in  the  units'  and  hundreds' 
places  interchanged  by  adding  792.  Find  the  number. 


SIMPLE    INDETERMINATE    EQUATIONS.  191 

26.  Some  men  earning  each  $2.50  a  day,  and  some  women 

earning  each  $1.75  a  day,  receive  all  together  for 
their  daily  wages  $44.75.  Determine  the  number 
of  men  and  the  number  of  women. 

27.  A  wishes  to  pay  B  a  debt  of  £1  12s.,  but  has  only  half- 

crowns  in  his  pocket,  while  B  has  only  4  penny-pieces. 
How  may  they  settle  the  matter  most  simply  ? 

28.  Show  that  323  a;  —  b21y  =  1000  cannot  be  satisfied  by 
.    integral  values  of  x  and  y. 

29.  A  farmer  buys  oxen,  sheep,  and  hens.      The  whole 

number  bought  is  100,  and  the  whole  price  £100. 
If  the  oxen  cost  £5,  the  sheep  £1,  and  the  hens  Is. 
each,  how  many  of  each  did  he  buy  ? 

30.  A  number  of  lengths  3  feet,  5  feet,  and  8  feet  are  cut ; 

how  may  48  of  them  be  taken  so  as  to  measure  175 
feet  all  together  ? 

31.  A  field  containing  an  integral  number  of  acres  less 

than  10  is  divided  into  8  lots  of  one  size,  and  7  of 
4  times  that  size,  and  has  also  a  road  passing  through 
it  containing  1300  square  yards.  Find  the  size  of 
the  lots  in  square  yards. 

32.  Two  wheels  are  to  be  made,  the  circumference  of  one 

of  which  is  to  be  a  multiple  of  the  other.  What  cir- 
cumferences may  be  taken  so  that  when  the  first  has 
gone  round  three  times  and  the  other  five,  the  differ- 
ence in  the  length  of  rope  coiled  on  them  may  be  17 
feet? 

33.  In  how  many  ways  can  a  person  pay  a  sum  of  £15  in 

half-crowns,  shillings,  and  sixpences,  so  that  the 
number  of  shillings  and  sixpences  together  shall  be 
equal  to  the  number  of  half-crowns  ? 


CHAPTER  XIV. 
INEQUALITIES. 

174.  Different  expressions  containing  any  given  letter 
will  have  their  values  changed  when  different  values  are 
assigned  to  that  letter ;  one  may  be  for  some  values  of  the 
letter  greater  than  the  other,  and  for  some  values  of  the 
letter  smaller  than  the  other. 

175.  Two  expressions,  however,  may  be  so  related  that, 
whatever  values  may  be  given  to  the  letter,  one  of  the 
expressions  cannot  be  greater  than  the  other. 

Thus,  2x'^  x^  +  1,  whatever  value  be  given  to  x. 
Note.  The  signs  5C  and  >•  are  read  not  less  than  and  not  greater 
than,  respectively. 

176.  For  finding  whether  this  relation  holds  between 
two  expressions,  the  following  is  a  fundamental  proposition : 

If  a  and  h  are  unequal,  a^  +  5^  >  2ah. 

For,  (a  —  Vf  must  be  positive,  whatever  the  values  of  a  and  h. 

That  is,  (a  -  hf  >  0, 

or  a2-2a6  +  62>o. 

.-.  a2  +  62>2a&. 

177.  The  principles  applied  to  the  solution  of  equations 
may  be  applied  to  inequalities,  except  that  if  each  side  of 
an  equality  have  its  sign  changed,  the  inequality  will  be 
reversed. 

Thus,  if  a  >  &  ;  then  —  a  <  —  &. 


INEQUALITIES.  193 

(1)  If  a  and  h  are  positive,  show  that  a^-\-h^>o?h-\-ah'^. 

We  shall  have  «»  +  6^  >  a%  +  aV, 

if  (dividing  each  side  by  a  +  6), 

a2  -  a&  +  6'  >  ah, 
a?  +  h''>2ah. 
But  this  is  true  (§  176).      .-.  a'  +  S^  >  a^S  +  a62. 

(2)  Show  that  o?  ^  h" -^  c"  >  ab  ^  ac -{-  he. 

Now,  a2  +  62  >  2  ah, 

a2  +  c2>2ac,  (§176) 

62  +  c2>26c. 
Adding,  2o?  +  262  +  2c2  >  2a6  +  2ac  +  26c. 

.-.  a2  +  62  +  c2  >  a6  +  ac  +  he. 

Exercise  7J 
Show  that,  the  letters  being  unequal  and  positive : 
1.    a^  +  35^>25(a  +  5).  2.    o?h -\- ah'>^a'h\ 

3.  (a^  +  6^)(a*  +  ^0>(«'  +  ^T 

4.  a'5  +  aV  +  a52  +  Vc  +  ac'^  +  ^'c^  >  6  a5^. 

5.  The  sum  of  any  fraction  and  its  reciprocal  >  2. 

6.  lia?=o^-\-h'^,  and  2/^  =  c^+<^,  xy  <^  ac-]-hd,  or  ad-^-bc. 

7.  ah-\-aci-bc<{a-\-b-cf-^{a-\'C--bf-\-{b^c-a)\ 

8.  Which  is  the  greater,  {o?  +  5^)((?2  +  d")  or  (ac  +  bdy  ? 

9.  Which  is  ihe  greater,  a*  —  b*  or  4a^(a  — 5)  when  a>b  ? 

10.  Which  is  the  greater,  \|t"  +  \/—  oi"  Va  +  V3  ? 

11.  Which  is  the  greater,     1"     or  -? 

2  a  +  o 

12.  Which  is  the  greater,  — +-^or-  +  -? 

0^     (T       ha 


CHAPTER  XV. 

INVOLUTION   AND   EVOLUTION. 

178.  Involution.  The  operation  of  raising  an  expression 
to  any  required  power  is  called  involution. 

Every  case  of  involution  is  merely  an  example  of  multi- 
plication, in  which  the  factors  are  equal. 

179.  Index  Law.     If  m  is  a  positive  integer,  by  definition 

a"^  —  aX  aX  a to  m  factors.  §  19 

Consequently,  if  m  and  n  are  both  positive  integers, 
{pJ'Y  =  a"  X  a"  X  a" to  m  factors. 

—  {aX  a to  n  factors)(a  X  a to  n  factors) 

taken  m  times 

—  aX  aX  a to  Tun  factors 

This  is  the  index  law  for  involution. 
Also, 

(a'")"  —  a'««  =  («")"»; 

{aby  =  abxab to  n  factors 

=  {aX  a to  n  factors)(5  X  b to  n  factors) 

180.  If  the  exponent  of  the  required  power  is  a  composite 
number,  the  exponent  may  be  resolved  into  prime  factors, 
the  power  denoted  by  one  of  these  factors  found,  and  the 
result  raised  to  a  power  denoted  by  another  factor  of  the 
exponent;  and  so  on.  Thus,  the  fourth  power  may  be 
obtained  by  taking  the  second  power  of  the  second  power ; 
the  sixth  by  taking  the  second  power  of  the  third  power. 


INVOLUTION   AND   EVOLUTION.  195 

181.  From  the  Law  of  Signs  in  multiplication  it  is  evi- 
dent that  all  even  powers  of  a  number  are  positive;  all  odd 
powers  of  a  number  have  the  same  sign  as  the  number  itself. 

Hence,  no  even  power  of  any  number  can  be  negative; 
and  the  even  powers  of  two  compound  expressions  which 
have  the  same  terms  with  opposite  signs  are  identical. 

Thus,         {h-ay^\-{a-h)Y^{a~h)\ 

182.  Binomials.     By  actual  multiplication  we  obtain, 

{a  +  by  =  a'-\-2ah  +  h''', 

(a  +  5)^  -  a^  +  3  a'h  +  3  a5^  +  5^ 

[a  +  by  =  a*  +  ^a'b  -\-^a'b''  +  4a6'  +  b'. 

In  these  results  it  will  be  observed  that : 

I.  The  number  of  terms  is  greater  by  one  than  the 
exponent  of  the  power  to  which  the  binomial  is  raised. 

II.  In  the  first  term,  the  exponent  of  a  is  the  same  as 
the  exponent  of  the  power  to  which  the  binomial  is  raised ; 
and  it  decreases  by  one  in  each  succeeding  term. 

III.  b  appears  in  the  second  term  with  1  for  an  exponent, 
and  its  exponent  increases  by  1  in  each  succeeding  term. 

IV.  The  coefficient  of  the  first  term  is  1. 

V.  The  coefficient  of  the  second  term  is  the  same  as  the 
exponent  of  the  power  to  which  the  binomial  is  raised. 

VI.  The  coefficient  of  each  succeeding  term  is  found 
from  the  next  preceding  term  by  multiplying  the  coefficient 
of  that  term  by  the  exponent  of  a,  and  dividing  the  product 
by  a  number  greater  by  one  than  the  exponent  of  b. 

If  b  is  negative,  the  terms  in  which  the  odd  powers  of  b 
occur  are  negative.     Thus, 

(a  -  by  =  a'-^a'b  +  6a'b'  -  ^a¥  +  b\ 

By  the  above  rules  any  power  of  a  binomial  of  the  form 
a  +  6,  or  a  —  6,  may  be  written  at  once. 


196  ALGEBRA. 

183.  The  same  method  may  be  employed  when  the  terms 
of  a  binomial  have  coefficients  or  exponents. 

(1)  {a-hf  =  a?  -^a^h  +  ?>ah''  -h\ 

(2)  (5a;2- 22/3)3, 

=  (5  x'f  -  3  (5  x^)\2y^)  +  3  (5  x''){2y^f  -  {2y^f, 
=  125  x^  -  150  xY  +  60  xY  -  8  y^. 

In  like  manner,  a  polynomial  of  three  or  more  terms 
may  be  raised  to  any  power  by  inclosing  its  terms  in  paren- 
theses, so  as  to  give  the  expression  the  form  of  a  binomial. 

(3)  (x3- 2x2  + 3a; +  4)2, 

=  {(x3-2a;2)  +  (3a;  +  4)}2, 

=  (3^-2x2)2  +  2{x^-2x^){3x  +  4)  +  {Sx  +  4)^, 

=  rr^  -  4x5  +  4a;4  +  6a;*  -  43^3  _  i6a;2  +  9x2  +  24a;  +  16, 

=  a;«  - 4a;5  +  lOx*  -  4x3  _  7,^.2  +  24x  +  16. 

Exercise  72. 
Perform  the  indicated  operations  : 


1.  (aj. 

11.  (2a^5c')'. 

21.  (-Sa'b'cf. 

2.  (^)'. 

12.  (-5ax'y')'. 

22.  (-3a;y7. 

3.  (Ty)^ 

13.  (-TTO'wxyy. 

23.  (-5a^^»a;7. 

4    /^«'*Y 

■A  2  J 

14.  f    2^Y. 
\     3  aocj 

-  e%f)' 

5.  f3^Y. 

15.  (3a;+l)\ 

-(-^T 

6.  (3:  +  2)'. 

16.  (2x-ay. 

26.  (1-a-  aj. 

7.(^-2)*. 

17.  (8a;  +  2a)^ 

27.  (2-3a;+4a:^y. 

8.(x  +  Sy. 

18.  (2a;-y)*. 

28.  {l-2x-\-xJ. 

9.  (1  +  2a;)'. 

19.  (x''y-2xyy. 

29.  (l-rr  +  a;^. 

10..(2ot-1)'. 

20.  (a6  -  3)^ 

30.  (l  +  a;  +  a^^)*. 

INVOLUTION    AND    EVOLUTION.  197 

184.  Evolution.  The  operation  of  finding  any  required 
root  of  an  expression  is  called  evolution. 

Every  case  of  evolution  is  merely  an  example  of  factor- 
ing, in  which  the  required  factors  are  all  equal.     Thus,  the 

square,  cube,  fourth roots  of  an  expression  are  found 

by  taking  one  of  the  two,  three,  four equal  factors  of  the 

expression. 

The  symbol  which  denotes  that  a  square  root  is  to  be 
extracted  is  ^\  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  written  above  to  indicate  the  root ; 
thus,  ■{/,  -y/,  etc.,  signifies  the  third  root,  fourth  root,  etc. 

185.  Index  Law.     If  m  and  w  are  positive  integers, 

{arf=a'^''.  §179 

Therefore  ^"aT"  =  a"^. 

Hence,  the  root  of  a  simple  expression  is  found  by  divid- 
ing the  exponent  of  each  factor  hy  the  index  of  the  root,  and 
taking  the  product  of  the  resulting  factors. 

Thus,  the  cuhe  root  of  a^  is  o? ;  the  fourth  root  of  81  a}^,  that  is, 
3*a^^  is  3a^;  and  so  on. 

The  above  is  the  index  law  for  evolution. 

Also,  since  {ahf  =  ft"5", 

therefore,  Va^  =-  a5  =  Va"  X  V^. 

186.  It  is  evident  from  §  181  that 

I.   Any  even  root  of  a  positive  number  will  have  the 
double  sign,  ±. 

II.    There  can  be  no  even  root  of  a  negative  number. 

III.  Any  odd  root  of  a  number  will  have  the  same  sign 
as  the  number. 

187.  The  indicated  even  root  of  a  negative  number  is 
called  an  impossible,  or  imaginary,  number. 


198  ALGEBRA. 

188.  if  the  root  of  a  number  expressed  in  figures  is  not 
readily  detected,  it  may  be  found  by  resolving  the  number 
into  its  prime  factors.  Thus,  to  find  the  square  root  of 
3,415,104 : 


2' 

3415104 

2' 

426888 

3^ 

53361 

7 

5929 

7 

847 

11 

121 

11 

3,415,104  =  2'' X  3'^  X  7'^X  IP. 


.-.  V3415104  =  2'x3  x7  xll  =  1848. 

Exercise  73. 
Simplify : 

1.  Va*,  Vx',  V4^,  ^64,  \/^Vy^,  -^/iQa^'b'c',  V-^2^'. 

2.  V-inSM'xY,  -v/3375F?^,  73111696 A*. 

3.  V5336UVyV«,^-^^3^.    ^^^, 

4.  V25^?F?+  V^^^¥c'-  a/81«*6V -  V^2a'b''c\ 


5.    <j21xY  X  V243yV  X  Vl6;rV. 

When  a  =  1,  5  =  3,  a:  =  2,  3/  =  6,  find  the  values  of: 


6.    4V2a;  —  ^abxy  -\-  b-\/a^h^xy. 


7.    2a-y/Sax^b-\/l2hy-{-^ahx-\/bxy. 


8.  -s/a^-\-2ab  +  b''X</a:'-\-2>a'b  +  ^ab'  +  b\ 

9.  Vb'  -Wa  -{-Ua^-a'-^  -y/b'  +  a'  -  2ab. 


INVOLUTION    AND    EVOLUTION.  199 

189.  Square  Roots  of  Compound  Expressions.  Since  the 
square  oi  a-j-b  is  a^  -\- 2  ab  -{-  b"^,  the  square  root  of 

It  is  required  to  devise  a  method  of  extracting  the  square 
root  a-\-b  when  a^  -\- 2  ab  -\- h"^  is  given. 

The  first  term,  a,  of  the  root  is  obviously  the  square  root  of  the 

first  term,  a\  in  the  expression. 

a^  +  2a&  +  b''\a±b_        if  the  a^  is  subtracted  from  the  given 

9l  expression,    the    remainder    is    2  a&  +  V^. 

2a  +  0       2ab  +  0^  Therefore,  the  second  term,  b,  of  the  root 

2ao  +  0^  is  obtained  by   dividing    the    first    term 

of   this    remainder    by    2  a,   that    is,   by 

double  the  part  of  the  root  already  found.      Also,  since   2ab  +  b'^ 

=  {2a  +  b)b,  the  divisor  is  completed  by  adding  to  the  trial-divisor  the 

new  term  of  the  root. 

The  same  method  will  apply  to  longer  expressions,  if  care 
be  taken  to  obtain  the  trial-divisor  at  each  stage  of  the 
process,  by  doubling  the  part  of  the  root  already  found,  and 
to  obtain  the  complete  divisor  by  annexing  the  new  term  of 
the  root  to  the  trial- divisor. 

Find  the  square  root  of 

1  +  10a;' -\-2bx'-{-  l^x^ -  2^x'> -  2Qx'  -  4:x. 


8x3 


16  a* 

-24x5  + 

25  x^ 

-20x3  +  10x2-4x  +  l|4r''- 

-3x2 +  2x- 

-1 

16  x« 
-3x2 

-24x'5  +  25x* 

-24x5+    9x^ 

-20x3  +  10x2 

8x3- 

6x2 +  2x 

16  X* 

16x*-12x3+    4x2 

0 

8a;3-6x2  +  4.'( 

:-  1 

-    8x3+    6x2-4x  +  l 

_    8x3+    6a;2_4  3,  +  | 

The  expression  is  arranged  according  to  descending  powers  of  x. 
It  will  be  noticed  that  each  successive  trial-divisor  may  be  obtained 
by  taking  the  preceding  complete  divisor  witli  its  last  term  doubled. 


200  ALGEBRA. 

Exercise  74. 
Extract  the  square  root  of : 

1.  a*  +  4a'-f-2a^-4a  +  l. 

2.  a:*-2:rV  +  3a;y-2:r/  +  ?/*. 

3.  4:a'—12a'x  +  6 aV  +  6 aV  +  a'x\ 

4.  9x^-12xy+16xy-24:xY-{-4:f-{-16xy\ 

5.  4a«+16c«+16aV-32aV. 

6.  4a;*4-9-30a;-20a;3  +  37a:^ 

7.  16a;*  -  IQahx'  +  16Z>V  +  4a^^^  ~  8ab'  +  45*. 

8.  a;'  +  25a;'  +  10a;*  -  4x'  -  20a;'  +  16  -  24a;. 

9.  x^  +  8a;*?/'  -  4a;V  —  4:xf  +  8a;y  —  10a;y  +  /. 

10.  4-12a-lla*  +  5a'-4a^  +  4a«+14al 

11.  9a''~6ab  +  S0ac  +  6ad+b''-10bc-2bd 

+  25c'+10cd+d\ 

12.  25a;«  -  31a;*y'  +  34a;y  -  30a;^2/  +  /  -  Sa;?/^  +  10a;y. 

13.  m'  — 4m^  +  10m«— 20m^-44m'4-35w* 

+  46m' -40m +  25. 

14.  a;*  —  a;'y  —  |-a;y  +  xt/^  +  y*. 

15.  2;*-4.^V  +  6:^y-6a;/  +  5y*-^  +  ^. 

17.  i+4_^104_20_^25_^24^16_ 

X      x^      x^       a;*      x^       x^ 

18.  «'_2^  +  3-2*  +  ^.       19.   ^•  +  ^'^5£:_2  +  i. 
&=       J  a      a'  ^  12      3     9 


INVOLUTION    AND    EVOLUTION.  201 

190.  Arithmetical  Square  Boots.  In  the  general  method 
of  extracting  the  square  root  of  a  number  expressed  by- 
figures,  the  first  step  is  to  divide  the  figures  into  groups. 

Since  1  =  1^,  100  =  10^,  10,000  =  100^,  and  so  on,  it  is  evident  that 
the  square  root  of  any  integral  square  number  between  1  and  100  lies 
between  1  and  10 ;  the  square  root  of  any  integral  square  number 
between  100  and  10,000  hes  between  10  and  100.  In  other  words, 
the  square  root  of  any  integral  square  number  expressed  by  one  or 
two  figures  is  a  number  of  one  figure ;  the  square  root  of  any  integral 
square  number  expressed  by  three  or  four  figures  is  a  number  of  two 
figures  ;  and  so  en. 

If,  therefore,  an  integral  square  number  be  divided  into  groups  of 
two  figures  each,  from  the  right  to  the  left,  the  number  of  figures  in 
the  root  will  be  equal  to  the  number  of  groups  of  figures.  The  last 
group  to  the  left  may  consist  of  only  one  figure. 

Find  the  square  root  of  3249. 

32  49(57  In  this  case,  a  in  the  typical  form  a^  +  2ab  +  b^ 

25 represents  5  tens,  that  is,  50,  and  b  represents  7. 

107)7  49  The  25  subtracted  is  really  2500,  that  is,  a^,  and  the 

7  49  complete  divisor  2a  +  6is2x50  +  7-  107. 

The  same  method  will  apply  to  numbers  of  more  than 
two  groups  by  considering  that  a  in  the  typical  form  repre- 
sents at  each  step  the  part  of  the  7'oot  already  found,  and 
that  a  represents  tens  with  reference  to  the  next  figure  of 
the  root. 

191.  If  the  square  root  of  a  number  have  decimal  places, 
the  number  itself  will  have  twice  as  many. 

Thus,  if  0.11  be  the  square  root  of  some  number,  the  number  will 
be  (0.11)2  ^  Qii  y^  Qii  _  0.0121.  Hence,  if  a  given  number  contains 
a  decimal,  we  divide  it  into  groups  of  two  figures  each,  by  beginning 
at  the  decimal  point  and  proceeding  toward  the  left  for  the  integral 
number,  and  toward  the  right  for  the  decimal.  We  must  be  careful 
to  have  the  last  group  on  the  right  of  the  decimal  point  contain  two 
figures,  annexing  a  cipher  when  necessary. 


202 


ALGEBRA. 


192.  If  a  number  contains  an  odd  number  of  decimal 
places,  or  gives  a  remainder  when  as  many  figures  in  the 
root  have  been  obtained  as  the  given  number  has  groups, 
then  its  exact  square  root  cannot  be  found.  We  may, 
however,  approximate  to  the  exact  root  as  near  as  we  please 
by  annexing  ciphers  and  continuing  the  operation. 

Find  the  square  roots  of  3  and  357.357. 

3.(1.732 •      3  57.35  70(18.903 

1 1 

27)2  00  28)2  57 
1  89  2  24 


343)1100  369)33  35 

10  29  33  21 


3462)  71  00  37803)  14  70  00 

69  24  1134  09 


193.  The  square  root  of  a  common  fraction  is  found  by 
extracting  the  square  root  of  the  numerator  and  of  the 
denominator.  But,  when  the  denominator  is  not  a  perfect 
square,  it  is  best  to  reduce  the  fraction  to  a  decimal  and 
then  extract  the  root. 

Exercise 

Extract  the  square  root  of: 

1.  120,409;  4816.36;  1867.1041;   1435.6521;  64.128064. 

2.  16,803.9369;  4.54499761;  0.24373969;  0.5687573056. 

3.  0.9;  6.21;  0.43;  0.00852;  17;  129;  347.259. 

4.  14,295.387;  2.5;  2000;  0.3;  0.03;  111. 

5.  0.00111;  0.004;  0.005;  2;  5;  3.25;  8.6. 

fi        1.16.100.     1 69  .     289  .     400 
"•     ^»    ¥"S''    T¥l^'    22  5'    ^2T>    TTS- 

'^'  i;  I;  i;  T2  5  rfc;  tIit'  f;  tV 


INVOLUTION    AND    EVOLUTION. 


194,   Oube  Boots  of  Compound  Expressions.     Since  the  cube 
of  a  -f  5  is  a'  +  3  a^6  +  3  alf  +  ¥,  the  cube  root  of 

a'  +  2>a^h  +  3a6'  +  6'  is  a  +  ^. 

It  is  required  to  devise  a  method  for  extracting  the  cube 
root  a-\-h  when  a^  +  3  a^b  +  3  al/  +  ¥  is  given. 


3a2 

+  3  a6  +  62 

a_3 

3a26  +  3a62  +  63 

3a2  +  3a6  +  62 

3  a26  +  3  a62  +  ^3 

The  first  term  of  the  root  is  a,  the  cube  root  of  a?. 

If  a?  is  subtracted,  the  remainder  is  ^a?h  +  Zah'^  +  6^;  therefore, 
the  second  term  h  of  the  root  is  obtained  by  dividing  the  first  term 
of  this  remainder  by  three  times  the  square  of  a. 

Also,  since  3a^b  +  S  ab^  +  b^  =  {Sa^  +  3ab  +  b"^)  b,  the  complete  divisor 
is  obtained  by  adding  Sab  +  b'^  to  the  trial-divisor  Sa"^. 

The  same  method  may  be  applied  to  longer  expressions 
by  considering  a  in  the  typical  form  to  represent  at  each 
stage  of  the  process  the  part  of  the  root  already  found. 


Find  the  cube  root  of  x^ 


(3  x^  —  x){—  x) 


3(a;2  -  xf  =  3a;*  -  Ga.-^  +  3a'2 


3x5  +  5a;'-3:r 
|a;2-a 


a;« 

3  a;*                   a^ 

-3ar'+     x" 

-Ss^^bT^-Sx-l 
-3a;5  +  5a,3 

3a;*-3x'+    x' 

-3a;5        4.3a;*-    :t? 

(3x2_3a;-l)(-l)^ 


3a;2  +  3.r  +  1 


6.t3 


+  3a;  +  1 


3a;*  +  6a;3_3a._i 


3.x"*  +  6ar»-3a;-l 


The  first  trial-divisor  is  3x*,  and  the  first  complete  divisor  is 
3  X*  —  3  a;^  +  a;^.  The  second  trial-divisor  is  3  {x^  —  xf,  or  3  a;*—  6  a;'+  3  x^. 
The  second  term  of  the  root  is  found  by  dividing  —  3a;^,  the  first  term 
of  the  remainder,  by  3  a;*,  the  first  term  of  the  root.  The  second 
complete  divisor  is  3x*  —  6a;'-f3a;-Fl. 


204  ALGEBRA. 

Exercise  76. 
Find  the  cube  root  of: 

1.  x^  +  ^xhj^\2xy''-^Sy\ 

2.  a' -  9a' +  27a -27. 

3.  a;' +  12a;' +  48:^  +  64. 

5.  a;'  +  3a;'  +  6:c*+7a;'  +  6a;'  +  3:^+l. 

6.  1  -  9a;  +  39a;'  -  99a;'  +  156a;*  -  144a;'  +  64.^;^ 

7.  a«-6a'  +  9a*  +  4a'-9a'-6a-l. 

8.  64a;«  +  192a;'  +  144a;*  -  32a;'  -  36a;'  +  12a;  -  1. 

9.  l-3a;+6a;'-10a;'  +  12a;*~12a;'+10a;'-6a;'+3a;'-a;^ 

10.  a«  +  9  a'b  -  135  o?b^  +  729  (i¥  -  729  b\ 

11.  c«  - 12  5c'  +  60  b'c'  - 160  b'c^  +  240  5 V  - 192  b'c  +  64  b\ 

12.  8a«+48a'6  +  60a*Z>'-80a'^/'-90a'6*+108a5'-275«. 

195.  Arithmetical  Oube  Eoots.  In  extracting  the  cube 
root  of  a  number  expressed  by  figures,  the  first  step  is  to 
divide  it  into  groups. 

Since  1  =  l^,  1000  =  10^,  1,000,000  =  lOO^,  and  so  on,  it  follows 
that  the  cube  root  of  any  integral  cube  number  between  1  and  1000, 
that  is,  of  any  integral  cube  number  which  has  one,  two,  or  three 
figures,  is  a  number  of  one  figure;  and  that  the  cube  root  of  any 
integral  cube  number  between  1000  and  1,000,000,  that  is,  of  any 
integral  cube  number  which  h&,s  four,  five,  or  six  figures,  is  a  number 
of  two  figures  ;  and  so  on. 

If,  therefore,  an  integral  cube  number  be  divided  into  groups  of 
three  figures  each,  from  right  to  left,  the  number  of  figures  in  the 
root  will  be  equal  to  the  number  of  groups.  The  last  group  to  the 
left  may  consist  of  one,  two,  or  three  figures. 

196.  If  the  cube  root  of  a  number  have  decimal  places, 
the  number  itself  will  have  thi^ee  times  as  many. 


INVOLUTION    AND    EVOLUTION. 


205 


Hence,  if  a  given  number  contains  a  decimal,  we  divide  the  figures 
of  the  number  into  groups  of  three  figures  each,  beginning  at  the  deci- 
mal-point and  proceeding  toward  the  left  for  the  integral  number, 
and  toward  the  right  for  the  decimal.  We  must  annex  ciphers  if  neces- 
sary, so  that  the  last  group  on  the  right  may  contain  three  figures. 

If  the  given  number  is  not  a  perfect  cube,  zeros  may  be  annexed, 
and  an  approximate  value  of  the  root  found. 

197.    In  the  typical  form,  the  first  complete  divisor  is 

the  second  trial-divisor  is  2>{a-\-bf,  or  Sa^  +  6a6 -f  3^'^ 
which  may  be  obtained  by  adding  to  the  preceding  com- 
plete divisor  its  second  term  and  twice  its  third  term. 
Extract  the  cube  root  of  5  to  five  places  of  decimals. 


5.000(1.70997 
1 

3  X  102  =  300 

4000 

3(10x7)  =210 

72=  49 -V 

559  [ 

3913 

259  J 

87000000 

3x17002  =  8670000 

3(1700x9)=  45900 

9'^=     81  ^ 

8715981  [ 

78443829 

45981  J 

85561710 

3X17092  =  8762043 

78858387 

67033230 

61334301 

After  the  first  two  figures  of  the  root  are  found,  the  next  trial- 
divisor  is  obtained  by  bringing  down  the  sura  of  the  210  and  49 
obtained  in  completing  the  preceding  divisor  ;  then  adding  the  three 
lines  connected  by  the  brace,  and  annexing  two  ciphers  to  the  result. 

The  last  two  figures  of  the  root  are  found  by  division.  The  rule 
in  such  cases  is,  that  two  less  than  the  number  of  figures  already 
obtained  may  be  found  without  error  by  division,  the  divisor  being 
three  times  the  square  of  the  part  of  the  root  already  found. 


206 


ALGEBRA. 


Exercise  77. 
Find  the  cube  root  of : 


1.  274,625. 

2.  110,592. 

3.  262,144. 

4.  884,736. 

5.  109,215,352. 

6.  1,481,544. 


7.  1601.613. 

8.  1,259,712. 

9.  2.803221. 

10.  7,077,888. 

11.  12.812904. 

12.  56.623104. 


13.  33,076.161. 

14.  102,503.232. 

15.  820.025856. 

16.  8653.002877. 

17.  1.371330631. 

18.  20,910.518875. 


19.    91.398648466125. 


20.    5.340104393239. 


21.    Find  to  four  figures  the  cube  roots  of  2.5  ;  0.2;  0.01 ; 
4;  0.4. 

198.  Since  the  fourth  power  is  the  square  of  the  square, 
and  the  sixth  power  the  square  of  the  cube  ;  the  fourth  root 
is  the  square  root  of  the  square  root,  and  the  sixth  root  is 
the  cube  root  of  the  square  root.  In  like  manner,  the  eighth, 
ninth,  twelfth roots  may  be  found. 

Exercise  78. 
Find  the  fourth  root  of: 
1.    81a*  -  540  a^6  +  1350  a'i^  -  1500  a5^  +  6255*. 

Find  the  sixth  root  of : 

3.  64  -  192a;+  240:r'  -  160a:'  +  60a;*  -  12a;'  +  x\ 

4.  729a;«-1458a;'+1215a;*-540a;'+135a;^-18a;+l. 

Find  the  eighth  root  of : 

5.  l-8y4-28y^-56/+702/*-563/'-l-28/--8?/'+y«. 


CHAPTER  XVI. 
THEORY  OF  EXPONENTS. 

199.  If  ?2  is  a  positive  integer,  we  have  defined  a"  to 
mean  the  product  obtained  by  taking  a  as  a  factor  n  times. 
Thus  o?  stands  for  a  X  a  X  a ;  b^  stands  for  b  X  b  xb  X  b. 

200.  From  this  definition  we  have  obtained  the  following 
laws  for  positive  and  integral  exponents : 


I. 

a™  X  a"  =  «'"+'*. 

II. 

(a«)«  ==  a"*". 

11. 

^  =  a"*-",  if  711  >  n, 

[V. 

«/^«.«  =.  a-. 

V. 

(abf  ^  a^'b"". 

201.  Since  by  the  definition  of  a"  the  exponent  n  denotes 
simply  repetitions  of  a  as  a  factor,  such  expressions  as  a^ 
and  a~^  have  no  meaning  whatever.  It  is  found  convenient, 
however,  to  extend  the  meaning  of  a"  so  as  to  include 
fractional  and  negative  values  of  n. 

202.  If  we  do  not  define  the  meaning  of  a"  when  n  is 
a  fraction  or  negative,  but  require  that  the  meaning  of  a** 
must  in  all  cases  be  such  that  the  fundamental  index  law 
shall  always  hold  true,  namely, 

.a"*  X  a"  =  a"'+'*, 

we  shall  find  that  this  condition  alone  will  be  sufficient  to 
define  the  meaning  of  a**  for  all  cases. 


208  ALGEBRA. 

203.   To  find  the  Meaning  of  a  Practional  Exponent. 
Assuming  the  index  law  to  hold  true  for  fractional  expo- 
nents, we  have 

a^  X  a^  =  a^    ^  =  a^  =  a, 

a^  Xa^  Xa^  =  a^+^+^  -=  a^  =  a, 

s.         a         3.         a         a.L3_i_ara  12 

a^Xa*  Xa^  X  a""  =  a^^"^'^'' =  a^  =  a\ 

i  1  i  +  - *° "  terms  ? 

a"  X  a'* to  w  factors  =  a"    "  =  a"--=  a, 

^  ^  ^  '^^^ ton  terms  "L» 

a'^Xa'' to  ?i  factors  =  a**    "  =  a  "  =•  a"*. 

That  is,    a^  is  one  of  the  two  equal  factors  of  a, 

aP  is  one  of  the  three  equal  factors  of  a, 

a'^  is  one  of  the  four  equal  factors  of  a^, 

1 
a"  is  one  of  the  n  equal  factors  of  a, 

a"  is  one  of  the  n  equal  factors  of  «*". 
Hence      a*  =  Va  ;  a*  =  Va  ! 

112 

Also,        a"  X  a**  X  a"  X to  m  factors, 

_ -| f- _ torn  terms  — 

^^n      n      n  ^  ^n  ^ 

m 

m 

The  meaning,  therefore,  of  a**,  where  m  and  n  are  posi- 
tive integers,  is,  the  wth  root  of  the  mth  power  of  a,  or  the 
mth  power  of  the  nth  root  of  a. 

Hence  the  numerator  of  a  fractional  exponent  indicates 
a  power,  and  the  denominator  a  root ;  and  the  result  is  the 
same  when  we  first  extract  the  root  and  raise  this  root  to 
the  required  power,  as  when  we  first  find  the  power  and 
extract  the  required  root  of  this  power. 


THEORY    OF   EXPONENTS.  209 

204.  To  find  the  Meaning  of  a^ 
By  the  index  law, 

a°  X  a"*  =  a""^*"  =  a™. 

.-.  a"  =  1,  whatever  the  value  of  a  is. 

205.  To  find  the  Meaning  of  a  Negative  Exponent. 

If  n  stands  for  a  positive  integer,  or  a  positive  fraction, 
we  have  by  the  index  law, 

a"  X  a-"  =  a"-"  =  aP. 
But  a"=:l. 

.-.  a"X  a~"=  1. 

That  is,  a"  and  a~"  are  reciprocals  of  each  other  (§  149), 
so  that  a~"  =  — ,  and  a"  = 


206.  Hence,  we  can  change  any  factor  from  the  numerator 
of  a  fraction  to  the  denominator,  or  from  the  denominator 
to  the  liumersitoY,  provided  we  change  the  sign  of  its  exponent. 

Thus  -^  may  be  written  ab'^c'^d'^,  or  — — — 


'cf        '  '        a-'b-'(^d' 


207.  We  have  now  assigned  definite  meanings  to  frac- 
tional and  negative  exponents,  meanings  obtained  by 
subjecting  them  to  the  fundamental  index  law  of  positive 
integral  exponents ;  and  we  will  now  show  that  Law  II., 
namely,  (a"*)**  =  a"*",  which  has  been  established  for  positive 
integral  exponents,  holds  true  for  fractional  and  negative 
exponents. 

(1)  If  71  is  a  positive  integer,  whatever  the  value  of  w, 
we  have 

{arx'Y  =  a'^xa"'Xam to  n  factors, 

__  Qm+m+tn to  n  terms 


210  ALGEBRA. 


(2)  If  ?2  is  a  positive  fraction  -L,  where  'p  and  q  are  posi- 
tive integers,  we  have  ^ 

p        

{oC^Y  =  {a^)  1  =  Via'^y  ?  203 

=  V^  (1) 

mp 

-  aT  g  203 

p 

niX- 

=  a     « 

(3)  If  7L  is  a  negative  integer,  and  equal  to  —p,  we  have 

(«•»}»  =  (a")-P  =  ^--L.  1205 

=  -^  (1) 

=  a-^p  I  205 

(4)  If  n  is  negative  and  equal  to  the  fraction  _-l,  where 
p  and  q  are  positive  integers,  we  have  ^ 


1 


i/a'^p. 

TOP 

1 

1 

1 


205 

203 

203 


(I) 
?205 


Hence,  (a"*)"  =^  a*"",  for  all  values  of  m  and  n. 


THEORY   OF   EXPONENTS.  211 

208.  In  like  manner  it  may  be  shown  that  all  the  index 
laws  of  positive  integral  exponents  apply  also  to  fractional, 
and  negative,  exponents. 

Exercise  79. 

Express  with  fractional  exponents  : 

1.  V?;  ■v'^;  (V^)^   Va'-,  -Va';  (V~ay;  </^' 

2.  v"^;   -v^^y?;   -^aFW]  bVa'hc'x\ 
Express  with  radical  signs  : 

3.  a^;  aH^  ]  4:r^2/"^;  3:^^-^"^. 
Express  with  positive  exponents  : 

Write  in  the  form  of  integral  expressions : 
Sx2/       z       a  .      e^    .  x~^ .  x'^ 

Z'     '    XY  '    be  '    a'b-'  '    y-'i'    yi' 

Simplify  : 

6.  J  Xa^;  b^  xb^;  c^  X  c^^ ;  d}>  X  d^^. 

7.  m^  X  m~^  ;  n*  X  n'^"^ ;  o^  X  or  ;  a°  X  a'K 

8.  a^  X  Va  ;  c~-  X  V^;  y*  X  ^'y  ;  ^'^  X  V:r-\  ^ 

9.  abh  X  arhc^  ;  Jbh''  X  ah~h^d. 

10.    x^y^z^  X  x~^y~h~^  ;  x^yh^  X  x~^y~h~K 


212  ALGEBRA. 

12.  a^-^a^;  c^  ^  c^  ]  n^^^^n^;  a^-^^^. 

13.  (a')^^{a')^;  (c~^)i ;  (m"^)* ;  (J)-';  {x^)i 

14.  (p~^y'^;  (q^y^;  (a;~V')~^;  («^  X  a^)~Tt. 

15.  (4a~^)-^  (27^>-y5.  (64c^«)-^;  (32 O^- 

209.  Compound  Expressions  having  fractional  or  negative 
exponents  are  multiplied  and  divided  the  same  as  com- 
pound expressions  having  positive  integral  exponents. 

(1)  Multiply  y^  +  y^  +  y^+1  by  y^-1. 


i-1 


y  +  y^  +  y^  +  y^ 
-y^-y^-y^-l 


y  —  1  y  —  I.  Ai 

(2)   Divide  x^  +  x^  -  12  by  x^  -  3. 


4x3-12 

4a;^  — 12  rc^  +  4.  ^?i«. 


Exercise  80. 
Multiply : 

1.  x""^  +  x^f -\- y""^  by  x""^  -  x^y^  ^y''^. 

2.  ru""*~"— 3/"  by  a:" +  2/""*"". 

3.  a:^  —  2x^  +  1  by  a;*  —  1. 

H  8a^  +  4a'7»^  +  5a^6^  +  9^>'  by  2a^-h^. 


THEORY   OF   EXPONENTS.  213 

5.  l  +  ah-'  +  a'b-'  by  1  -  ab-' +  a'b-\ 

6.  a'b''  +  2  +  a-'b'  by  a'b-'  -  2  -  a''b\ 

7.  4:x  -^  -i-Sx-^-\-  2x-'  +  1  by  a;"'  -  rr"^  +  1. 
Divide : 

8.  a;*"  —  ?/*"  by  a;"  —  2/^ 

9.  a:  +  2/  +  z  — 3a;*yy  by  x^  +  y^-j-z^. 

10.  37  +  2/  by  x'^  —  a;^2/^  +  ^^2/^  ~  ^2/^  +  2/^- 

11.  x'^y~'^  +  2  +  a;~y  by  a;?/"^  +  ^~V- 

12.  a-*  +  a-'b-'  +  5-*  by  a"^  -  a'^b-'  +  ^>-^ 
Find  the  squares  of : 

13.  4a5-^  J-b^;  a  +  a-';  2ah^-arh^. 

If  a  =  4,  5  —  2,  c  =^  1,  find  the  values  of : 

14.  ah]  bab-')  2{abf',  otH-'c^ \  l2a-'b-\ 

15.  Expand  (a*  -  b~^y  ;  (2a;-^  ■\- x)' ;  (a^-^  -  ^y-^'. 
Extract  the  square  root  of : 

16.  9a;-*  -  ISa;-^^  +  ISa;-^  -  Sa;-^*  +  y\ 

Extract  the  cube  root  of : 

17.  8ar»  +  12a;^  -  30a;  ~  35  +  45a;-'  +  27 a;"^  -  27 a;-^ 
Eesolve  into  prime  factors  with  fractional  exponents  : 

18.  a/12,  a/72,  \/96,  ^64;  and  find  their  product. 
Simplify : 

19.  S(^"")'X(^'')~'P"^-         20.    (x'^'' X  x~^''f^\ 

21 .  3  (a^  +  b^^y  -  4  (a^  +  &*)(«*  -  b^)  +  («^  -  2 b^J. 

22.  Ka'^f-^I'^T.  24.    [ K«"")~"K]'-^[K«")"i ""]"'• 


CHAPTER  XVII. 

RADICAL   EXPRESSIONS. 

210.  A  radical  expression  is  an  expression  affected  with 
the  radical  sign ;  as,  Va,  V9,  Va^  Va  +  b,  V32. 

211.  An  indicated  root  that  cannot  be  exactly  obtained 
is  called  a  surd,  or  irrational  number.  An  indicated  root  that 
can  be  exactly  obtained  is  said  to  have  the /orm  of  a  surd. 

The  required  root  shows  the  order  of  a  surd ;  and  surds 
are  named  quadratic,  cubic,  biquadratic,  according  as  the 
second,  third,  or  fourth  roots  are  required. 

The  product  of  a  rational  factor  and  a  surd  factor  is 
called  a  mixed  surd;  as  3V2,  &Va.  The  rational  factor 
of  a  mixed  surd  is  called  the  coefficient  of  the  radical. 

When  there  is  no  rational  factor  outside  of  the  radical 
sign,  that  is,  when  the  coefficient  is  1,  the  surd  is  said  to 
be  entire ;  as,  V2,  Va. 

212.  A  surd  is  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  integral  and  as  small  as  possible. 

Surds  which  have  the  same  surd  factor,  when  reduced 
to  the  simplest  form,  are  said  to  be  similar. 

Note.  In  operations  with  surds,  arithmetical  numbers  contained 
in  the  surds  should  be  expressed  in  their  prime  factors. 

Reduction  of  Radicals. 

213.  To  reduce  a  radical  is  to  change  its /orm  without 
changing  its  value. 


RADICAL    EXPRESSIONS.  215 

Case  I. 

214.  When  the  Kadical  is  a  Perfect  Power  and  has  for  an 
Exponent  a  Factor  of  the  Index  of  the  Eoot. 

(1)  A/^=a^-a^=V^j^  ^ 

(2)  A/36^  =  \/(6^^(6a^))^  =  (6a5)^-V6^; 

(3)  •\/25a*6V«  =  -s/{bdWy  =  (5 a'bc'f  -■=  (5 a'bc'f 

We  have,  therefore,  the  following  rule : 

Divide  the  index  of  the  root  by  the  exponent  of  the  power. 


Simplify : 

1.  </36. 

2.  ^8l. 

Exercise  81. 

6.  V^a'b\ 

7.  V^a'b\ 

11. 
12. 
13. 

6/36  c'' 
\49a^ 

Ax -by 
^{x-\-2>y 

3.    ^125. 

8.    Vl6a*6\ 

4.    \/100. 

9.    VM2>a'b\ 
10.    ^/81a*6^ 

6  %a'b^ 

5.    -^343. 

\27ry 

Case  II. 


215.   When  the  Kadical  is  the  Product  of  Two  Factors,  One  of 
which  is  a  Perfect  Power  of  the  Same  Degree  as  the  Eadical. 

Since  V^b  =  V^  X  V^  =  a^b  (§  185),  we  have 

(1)  Va^  =  Vo"  X  V6  ==:  a  V6  ; 

(2)  -v/I08  =  a/27x4  =  ^/27x^/4  =  3^; 


216 


ALGEBRA. 


(3)  4V72a2/^='  =  WSQd'b'  X  26  =.  WSQa'b'  x  V26 
=  4  X  6abV2b  =  24a^/V2l; 


(4)  2V54a*6  =  2V27a'  X  2ab  =  2</21  a'  x  V2^ 
=  2  X  3aA/2^  =  6a-\/2^. 

We  have,  therefore,  the  following  rule : 

Resolve  the  radical  into  two  factors,  one  of  which  is  the 
greatest  perfect  power  of  the  same  degree  as  the  radical. 

Remove  this  factor  from  under  the  radical  sign,  extract 
the  required  root,  and  multiply  the  coefficient  of  the  radical 
by  the  root  obtained. 


Simplify : 

1.  Vl25. 

2.  V243. 

3.  \/l62. 

4.  a/256. 

5.  ^375. 

6.  \/320. 

7.  a/'486. 

8.  a/729. 

9.  -v/208. 

10.  V605. 

11.  2^/144. 


Exercise  82. 

13.  7^/176. 

14.  l^rri^n. 

15.  5\/W. 

16.  6-v^^^. 

17.  3-^^^». 

18.  7\/64^. 

19.  6^/108 mV 


25 


26 


'l27a^ 
\64a;y' 


27. 


49cW 
'36:ry 

\'l25/ 


28 


•    ^  SI  3125' 


20.  4vvy'- 


12.    3V2662. 


21.  2V-1029. 

22.  -y/-  1458. 

23.  3\/l875. 

24.  4\/ 


-•  ^\S- 


30.    2 


3p-  -  y)-V 

3j     4a^     /75?J 
ScViea^^*^' 


RADICAL    EXPRESSIONS.  217 


Case  III. 

216.  When  the  Kadical  Expression  is  a  Fraction,  the  Denomi- 
nator of  which  is  not  a  Perfect  Power  of  the  Same  Degree  as 
the  Eadical. 


27x8 


I__  ^60  =  4v^60. 


3X2 

We  have,  therefore,  the  following  rule  : 

Multiply  both  terms  of  the  fraction  by  such  a  number  as 
will  make  the  denominator  a  perfect  power  of  the  same 
degree  as  the  radical;  and  then  proceed  as  in  Case  II. 

Exercise  83. 
Simplify : 

1.  VJ.  4.    3V|.  7.    VI.  10.    2^/S. 

2.  Vi  5.    2</S.  8.    ^S-  11-    S^S- 

3.  VJ.  6.    3VA-  9-    </S^-  12.    2</A. 


218  ALGEBRA. 

Case  IV. 
217.   To  reduce  a  Mixed  Surd  to  an  Entire  Surd. 
Since  aVb  =  ^ cC  X  Vh  =  Va"6,  we  have 
(1)  3  V5  -  V3^^^^  =  V9">r5  =  V45  ; 


(2)  a^h^bc  =  ^{a'hy  Xbc^-  ^a'b'  X  he  =  V^^  ; 

(3)  ^xV^  =  Mi^^x^Xxy  =  </8x'Xxy  -  a/8^  ; 

(4)  3y^-\/?  =  </(^yyxx'  =  ^/STf^. 

We  have,  therefore,  the  following  rule : 

liaise  the  eoefficient  to  a  power  of  the  same  degree  as  the 
radical,  multiply  this  power  hy  the  given  surd  factor,  and 
indicate  the  required  root  of  the  product. 

Exercise  84. 
Express  as  entire  surds  : 

1.  3V5.         5.    2VI.         9.    ~2Vy.  13.        ^Va. 

2.  3V2I.       6.    3^3.       10.    -3^7.  14.  -i^^. 

3.  3^2.         7.    2a/5.       11.    -mVlO.  15.  |Vm^ 

4.  2V'5.         8.    2</2.       12.    -2^x,  16.  ~lVm\ 


Case  V. 
218.   To  reduce  Eadicals  to  a  Common  Index. 
(1)  Reduce  V2  and  VS  to  a  common  index.  _.  Ihyr^ 

v^  ==  3^  =  3^  =  \/32  =  v^. 
We  have,  therefore,  the  following  rule : 


RADICAL   EXPRESSIONS.  219 

Write  the  radicals  with  fractional  exponents,  and  change 
these  fractional  exponents  to  equivalent  expo7ients  having  the 
least  common  denominator.  Baise  each  radical  to  the  power 
denoted  hy  the  numerator,  and  indicate  the  root  denoted  hy 
the  common  denominator. 

Exercise  85. 
Reduce  to  surds  of  the  same  order : 

^  1.  <Jl  and  ■^/2.  7.  V%  ^'3,  and  ^5. 

'-  2.  a/7  and  V6.  8.  </^\  </b,  and  </c. 

"3.  V3  and  a/4.  .         9.  </a\  a/?,  and  </I\ 

4.  Va  and  \^b'\  10.  ^x'^y,  'vabc,  and  V2z. 

5.  V5  and  VSS.  11.  Va;  —  3/ and  Va;  +  2/- 

6.  3^,  3^,  and  3i  12.  Va  +  b  and  \/a-b. 

Note.   Surds  of  diflferent  orders  ma; 
aie  order  and  then  compared  in  respe( 

Arrange  in  order  of  magnitude  : 

13.    2V3,  3V2,  fVi.  15.    2V22,  3V7,  4V2. 

14.  Vf ,  V{i.  16.  3VI9, 5V2, 3V3. 

Addition  and  Subtraction  of  Radicals. 

219.  In  the  addition  of  surds,  each  surd  must  be  reduced 
to  its  simplest  form  ;  and,  if  the  resulting  surds  are  similar, 

Find  the  algebraic  sum  of  the  coefficients,  and  to  this  sum 
annex  the  common  surd  factor. 

If  the  resulting  surds  are  not  similar, 
Connect  them,  with  their  proper  signs. 


Note.   Surds  of  different  orders  may  be  reduced  to  surds  of  the 
same  order  and  then  compared  in  respect  to  magnitude. 


220  ALGEBRA. 

(1)   Simplify  V27+  Vi8  +  Vll?. 

V27  =  (32  X  3f  =  3  X  3^  =  3  V3  ; 

V-i8  -  (2*  X  3)^  =  22  X  3^  =  4  X  3^  =  4V3 ; 

VI47  =  (72  X  3f  =  7  X  3^  =  7  V3. 

V27  +  V48  +  \/l47  =  (3  +  4  +  7)  V3  =  14\/3. 

•    (2)  Simplify  2^320-3^40. 

2v^320  =  2(26  X  5)^  -  2  X  22  X  5^  =  8^5  ; 
3  v^40  =  3  (23  X  5)3  =  3  X  2  X  5^  =  6  v'g. 
.-.  2v'320-3v/iO-(8-6)-^5  =  2v'^. 

_  (3)   Simplify  2V|  -  3 V|  +  V^, 


2Vf  =  2V-V  =  2\/15  X  i  =  I  Vl5 


3  Vf  =  3  V|f  -  3  VlSx^  =  |Vl5 


.-.  2 Vf  -  3  Vf  +  vs  =  (I  -  f  +  ^)v'i5  =  ^ vr5. 

Exercise  86. 
Simplify : 

1.  8vn  +  7vn-iovii. 

2.    3V5-5V5-f  7V5. 

3.  V2T  +  2V48  +  3VI08.       7.    VIOOO -f  V50  +  V288. 

4.  ^I28+ a/686  +  a/16.         8.    "v'^M  +  3a/I6  +  ^4^. 

5.  12V72  -  3vT28.  9.    7\/8l  -  3\/l029. 

6.  2 V3 -f  3Vli  -  V5|.         10.    Vf+V60-Vl5-V|. 


RADICAL   EXPRESSIONS.  221 

12.  Vf+vs-v^. 


13.  V4:a'b-^-y/2bab'-(a-bb)Vab. 

14.  c</W?-a-\/aF?+b-\/^^\ 

15.  2a/40  +  3\/108  +  -v/500- a/320 -2^1372. 

16.  V3"63-2V243  +  VT08-2Vl47. 

17.  a/189  -  2^/448  +  a/875  +  ^15l2. 

18.  a/T62- a/512 +  2a/32- a/1250. 

19.  a/^^^-3a/=^  +  2a/I92. 

20.  a/20  +  a/45 -a/|. 

21.  2V'c(^'  +  V8¥-\\j' 

22.  a/50  +  |V288 K. ^• 

a/2      a/450 

23.  Vl70l  +  iVSi  -  i V525. 

Multiplication  of  Radicals. 

220.   Since  Va  X  a/6  =  -y/ab,  we  have 

(1)  3a/8  X  5a/2  =  3x5xa/8xV2  =  15a/I6  =  60 ; 

(2)  3a/2  X  4a/3  =  3a^  X  4a/9  =  12a/72. 
We  have,  therefore,  the  following  rule  : 

Express  the  radicals  with  a  common  index.  Find  the 
product  of  the  coefficients  for  the  required  coefficient,  and  the 
product  of  the  surd  factors  for  the  required  surd  factor. 

Reduce  the  result  to  its  simplest  form. 


222  ALGEBRA. 

Exercise  87. 
Find  the  product  of : 

1.  3V2  by  4V6.  •        6.  -^VIO  by  j\\/T5. 

2.  2V5by3Vl5.    •  7.  5V|  by  f Vl62. 

3.  2Vl0by5Vl4.  8.  f V2l  by  y^-V^. 

4.  3V27  by  7 V48.  9.  ^I08  by  5V32. 

5.  2-^4by'6A/32.  10.  5^54  by  7V48. 

221.    Compound  radicals  are  multiplied  as  follows 
Multiply  2V3  +  3 V^  by  3V3  -  Wx. 

3V3-4\/x 
18  +  9\/3^ 

-SVSx-Ux 


18+    V3aj-12a; 


Exercise  88. 
Find  the  product  of : 

1.  (2V^-7)x3V^.  6.   (V2  4-V3-V5)l 

2.  (Va- V6)l  7.    (V5  +  3V2+V7)^ 

3.  (3V5-7V2)^  8.   (2V5-V2-V7)^ 


4.  (V7  +  3V3)(V7-2V3).   9.   (2Vx+-VS-2xy. 

5.  (3V5-V2)(V5-3V2).    10.    (2V^H^-3Vc?=y7. 


RADICAL   EXPRESSIONS.  223 

Division  of  Radicals. 

222.  Since  ^  =  i^iiL^?  =  yi,  we  have 

a  Va 

(1)  iV8=.2V4  =  4; 
2V2 

(2)  4^3  .^  ^W  _  4  v^3^x2^  _  ^/Tj^ 
2V2     2^2^         2a/^ 

We  have,  therefore,  the  following  rule  : 

Express  the  radicals  with  a  common  index.  Find  the 
quotient  of  the  coefficients  for  the  required  coefficient,  and 
the  quotient  of  the  surd  factors  for  the  required  surd  factor. 

Reduce  the  result  to  its  simplest  form. 

Exercise  89. 
Divide : 

1.  Vl62  by  V2.         4.    Vi  by  Vf  7.    V5  by  a/I. 

2.  \/8l  by  Vl.  5.    VS  by  Vf.         8.    V|  by  "V^J. 

3.  -s/tc^'  by  Va^       6.    2VJ  by  | V|.       9.    V^f  by  Vf . 

10.  3  V2  +  V72  -  3  V8  by  V3. 

11.  9V2-6V6-3V8  4-12V32by3V2. 

12.  V2-Vl  +  V\^-VUhy  V2. 

223.  The  quotient  of  one  surd  by  another  may  be  found 
by  rationalizing  the  divisor;  that  is,  by  multiplying  the 
dividend  and  divisor  by  a  factor  which  will  free  the  divisor 
of  surds. 

224.  This  method  is  of  great  utility  when  w^e  wish  to 
find  the  approximate  numerical  value  of  the  quotient  of 
two  simple  surds  ;  and  is  the  method  required  when  the 
divisor  is  a  compound  surd. 


224  ALGEBRA. 


(1)  Given  V2  =  1.41421,  find  the  value  of  ~ 

V2 
5  5V2         5V2     7.07105^3  53553 


V2      V2xV2        2  2 

(2)  Divide  3 V5  -  4 V2  by  2V5  +  3 V2. 

3v'5-4V2  _  (3V5-4V2)(2V5-3\/2)  _  54-17V10 
2V5  +  3V'2     (2V5  +  3V2)(2\/5-3\/2)         20-18 

^54-17Vip^27-.8^VIO. 

By  two  operations  the  divisor  may  be  rationalized  when 
it  consists  of  th7'ee  quadratic  surds. 

Thus,  if  V6+VS-V2  be  multiplied  by  V6  -_V3  +  a/2,  the 
result  will  be  6  -  5  +  2V6  =  1  +  2\/6  ;  and  if  1  +  2 V6  be  multiplied 
by  1  -  2  V6,  the  product  will  be  1  -  24  =  -  23. 


Exercise  90. 

Find  the  approximate 

value  of: 

1.     2.           2.      1. 

V3                  V5 

3     7V2             ^     2V5 
V3                   3V2 

Divide : 

^.    3by  V7  + V5. 

10.  7  +  2VlOby  7--2V10. 

6.    7by2V5-V6. 

11.  V5  -  V6  by  2  V5  -  V6. 

7.    6by5-2V6. 

12.  a-{-bhy  a—  Vb. 

8.    4  ->  V2  by  1  +  V2. 

13.  1  by  V5  4- V3  + V7. 

9.    V5+ V2  by  V'5  - V2.    14.  2  by  V5  -  5V2  -f  V7. 


radical  expressions.  225 

Involution  and  Evolution  of  Radicals. 

225.    Any  power  or  root  of  a  radical  is  easily  found  by 
using  fractional  exponents. 
(1)*  Find  the  square  of  2\/a. 

(2  y/af  =  (2  Jf  =  2^  a^  =  4  J  =  4  y/a\ 

(2)  Find  the  cube  of  2Va. 

(3)  Find  the  square  root  of  4  :rVa*6^ 

{^xV¥b^)^  =  {4:xah^)^  =  ^hhh^  =  4^  AM  =  2  v^^a^V: 

(4)  Find  the  cube  root  oi  4:X^a^b^. 


Exercise  91. 
Perform  the  operations  indicated  : 

1.  (a/8)*.  4.    (aVay,  7.    (^256)^. 

2.  (^64)3.  5.    (cc</-.y.  8.    g^l)^ 

3.  (Vly.  6.    (3^3)1  9.    (2^/47^)^. 

Properties  of  Quadratic  Surds. 

226.  The  product  or  quotient  of  two  dissi7nilar  quadratic 
surds  will  be  a  quadratic  surd.     Thus, 

Vai  X  '\/abc  =  ab^ c  \  ■\/abc  -f-  Va6  =  Ve. 

For  every  quadratic  surd,  when  simplified,  will  have 
under  the  radical  sign  one  or  more  factors  raised  only  to 
the  first  power  ;  and  two  surds  which  are  dissimilar  cannot 
have  all  these  factors  alike. 

Hence,  their  product  or  quotient  will  have  at  least  one 
factor  raised  only  to  the  first  power,  and  will  therefore  be 
a  surd. 


226  ALGEBRA. 

227.  The  sum  or  difference  of  two  dissimilar  quadratic 
surds  cannot  he  a  rational  number,  nor  can  it  he  expressed 
as  a  single  surd. 

For  if  Va  ±  V3  could  equal  a  rational  number  c,  we 
should  have,  by  squaring, 

a  zh  2Va6  +  ^  --  ^' ; 
that  is,  ±  2^ ah  =  c'^  —  a  —  h. 

Now,  as  the  right  side  of  this  equation  is  rational,  the 
left  side  would  be  rational ;  but,  by  §  226,  Vai  cannot  be 
rational.     Therefore,  Va  ±  V^  cannot  be  rational. 

In  like  manner,  it  may  be  shown  that  Va  ±  V^  cannot 
be  expressed  as  a  single  surd  V^. 

228.  A  quadratic  surd  cannot  equal  the  su7n  of  a  rational 
numher  and  a  surd. 

For  if  Va  could  equal  c  +  V^,  we  should  have,  by 
squaring, 

a  =  c''-{-2cVh  +  h, 

and,  by  transposing, 

2cVh^a-h-c\ 

That  is,  a  surd  equal  to  a  rational  number,  which  is 
impossible. 

229.  If  a-\-  V^-  =  x-\-  Vy,  then  a  will  equal  x,  and  h 
will  equal  y. 

For,  by  transposing,  V^  —  Vy  —  x  —  a\  and  if  h  were 
not  equal  to  y,  the  difference  of  two  unequal  surds  would 
be  -rational,  which  by  §  227  is  impossible. 

.',  h  ^=^y,  and  a  =  x. 

In  like  manner,  if  a~^h  =  x  —  ->/y,  a  will  equal  x, 
and  h  will  equal  y. 


RADICAL   EXPRESSIONS.  227 

230.  To  extract  the  Square  Eoot  of  a  Binomial  Surd. 
Extract  the  square  root  of  a  +  V^. 


Suppose                         V  a  +  Vft  =  Va;  +  y/y.  (1) 

By  squaring,                    a  +  V6  =  .r  +  2y/xy  +  y.  (2) 

■    a  =  x-\-y  and  V6  =  2\/xy.  §  229 

Therefore,                         a  —  V6  =  .t  -  2\/xy  +  y,  (3) 


and  Va  -  Vb  =  Vx  -  y/y.  (4) 

Multiplying  (1)  by  (4), 

Vd^  —  b  =  x  —  y. 
But  a  =  rr  +  y. 

a  +  y/a^  —  b 


Adding,  and  dividing  by  2,      x 
Subtracting,  and  dividing  by  2, 


a  —  V  a'^ 
2/  = iT 


From  these  two  values  of  x  and  y,  it  is  evident  that  this 
method  is  practicable  only  when  a^  -  6  is  a  perfect  square. 

(1)  Extract  the  square  root  of  7  +  4  V3. 


Let 

V^  +  Vy  =  V?  +  4\/3. 

Then 

V.r-  V^  =  V7-4V3. 

Multiplying, 

x-y  =  V49  - 48. 

.'.  re  -  3/  =  1. 

But 

x  +  y  =  1. 

.-.  a;  =  4,  and  y  =  3. 

.-.  VS  +  Vy  =  2  +  V3. 

V7  +  4V3  =  2  +  Vs. 


228  ALGEBRA. 

A  root  may  often  be  obtained  by  inspection.  For  this  purpose, 
write  the  given  expression  in  the  form  a  +  2y/h,  and  determine  what 
two  numbers  have  their  sum  equal  a,  and  their  product  equal  h. 

(2)  Find  by  inspection  the  square  root  of  18  +  2  V77. 

It  is  required  to  find  two  numbers  whose  sum  is  18  and  whose 
product  is  77;  and  these  are  evidently  11  and  7. 

Then        18  +  2\/77  =  11  +  7  +  2\/irxT,  . 
=  (\/rr+V^)2. 

That  is,    \/n  + V7  =  square  root  ©f  18  +  2  V77. 

(3)  Find  by  inspection  the  square  root  of  75  —  12V21. 

It  is  necessary  that  the  coefiicient  of  the  surd   be  2;  therefore, 
75  —  12\/21  must  be  put  in  the  form 

75  -  2  V756. 

The  two  numbers  whose  sum  is  75  and  whose  product  is  756  are 
63  and  12. 

Then  75  -  2  V756  -  63  +  12  -  2\/6lxT2. 

=  (^63  -  \/i2)2  =  (3V7  -  2>/3)2. 

That  is,       3  V7  -  2V3  =  square  root  of  75  -  12\/2l. 


Exercise  92. 
Extract  the  square  roots  of : 

1.  14  +  6V5.          6.    20-8V6.  11.  14  -  4 V6. 

2.  17 +  4 Vis.         7.    9-6V2.  12.  38  -  12 VlO. 

3.  IO  +  2V2I.         8.    94~42V5.  13.  103-12VlT. 

4.  16  +  2V55.         9.    13-2V30.  14.  57  -  12VI5. 

5.  9-2VT4.         10.    II-6V2.  15.  3I--V10. 

16.    2a4-2Va'-^>'.         18.    87  -  12 V42. 


17.    a^--2h^a'-h\        19.    {a  +  by-4:{a~h)y/ab. 


CHAPTER  XVIII. 
IMAGINARY   EXPRESSIONS. 

231.  An  imaginary  expression  is  any  expression  which 
involves  the  indicated  even  root  of  a  negative  number. 

It  will  be  shown  hereafter  that  any  indicated  even  root 
of  a  negative  number  may  be  made  to  assume  a  form  which 
involves  only  an  indicated  square  root  of  a  negative  num- 
ber. In  considering  imaginary  expressions,  we  accordingly 
need  consider  only  expressions  which  involve  the  indicated 
square  roots  of  negative  numbers. 

Imaginary  expressions  are  also  called  imaginary  numbers 
and  complex  numbers.  In  distinction  from  imaginary  num- 
bers, all  other  numbers  are  called  real  numbers. 

232.  Imaginary  Square  Eoots.  If  a  and  h  are  both  posi- 
tive, we  have 

I.    V^  -  Va  X  V^.         II.    ( Va)^  =  a. 

If  one  of  the  two  numbers  a  and  h  is  positive  and  the 
other  negative,  law  I.  is  assumed  still  to  apply ;  we  have, 
accordingly : 


V^  -  V4(^  =.  Vi  X  V-  1  -  2V-  1 ; 

V=^  =  V5(^  =  V5  X  V^^  ==  5 V=^; 

V^  =  V<=T)  =  Va  X  V^  =  a  V=l ; 
and  so  on. 

It  appears,  then,  that  every  imaginary  square  root  can 
be  made  to  assume  the  form  aV—  1,  where  a  is  a  real 
number. 


230  ALGEBRA. 

233.  The  symbol  V—  1  is  called  the  imaginary  unit,  and 
may  be  defined  as  an  expression  the  square  of  which  is  ~  1. 

Hence,    V^^  X  V^^  =  ( V^H^)'  =  -  1 ; 

V^  X  V^  =-Vax  V^  x-Vbx  V^ 

-  V^  X  (- 1) 

-  -  V^. 

234.  It  will  be  useful  to  form  the  successive  powers  of 
the  imaginary  unit. 

(V="l)  . -^  +  V^; 

i^^y =-1; 

( V^)^  -  ( V^)^  V^     =  (-  1)  V^  =  -  V=l ; 
( V~iy  =  ( V^)-'  (V^7  =  (-  l)(-  1)  =  + 1 ; 

( v=T)^  =  ( v=n:)*  v=n[   =  (+ 1)  v^=t =  +  v=~i ; 

and  so  on.     We  have,  therefore,  if  n  is  any  integer, 

235.  Every  imaginary  expression  may  be  made  to  assume 
the  form  a  -\-  h-y/—  1,  where  a  and  b  are  real  numbers,  and 
may  be  integers,  fractions,  or  surds. 

If  5  =  0,  the  expression  consists  of  only  the  real  part  a, 
and  is  therefore  real. 

If  rt  =  0,  the  expression  consists  of  only  the  imaginary 
part  Z>  V—  1,  and  is  called  a  pure  imaginary. 


IMAGINARY    EXPRESSIONS.  231 

236.  The  form  a-\-bV—  1  is  the  typical  form  of  imagi- 
nary expressions. 

Reduce  to  the  typical  form  6  +  V-  8. 
This  may  be  written  G+VSx  V^,  or  6  +  2V2X  V^; 
here  a  =  6,  and  b  =  2V2. 

237.  Two  expressions  of  the  form  a-\-b  V- 1,  a  —  b^—  1, 
are  called  conjugate  imaginaries. 

To  find  the  sum  and  product  of  two  conjugate  imagina- 
ries, 

a  +  ^V^ 
a-b^^^ 

The  sum  is  2  a 

a  -\-b  V^ 
a  -b  V^ 

a'  +  abV^ 
-abV^^  +  b' 

The  product  is        d^  +  b^ 

From  the  above  it  appears  that  the  sum  and  product  of 
two  conjugate  imaginaries  are  both  real. 

238.  An  imaginary  expression  cannot  be  equal  to  a  real 
number. 

For,  if  possible,  let 

a-{-b^/^^  =  c. 

Then  transposing  a,       5  V—  1  =  c  —  a, 
and  squaring,  —b'^  =  (c  —  a)^ 

Since  b^  and  (c  —  a)'  are  both  positive,  we  have  a  nega- 
tive number  equal  to  a  positive  number,  which  is  impossible. 


232  ALGEBRA. 

239.  If  two  imaginary  expressions   are  equal,  the  real 
parts  are  equal  and  the  imaginary  parts  are  equal. 

For  let  a  +  ^ V^  =  c  +  <iV^. 

Then  {h  -  d)^'^l=  c- a] 

squaring,  —  (jb  —  dy  =  {c  —  ay, 

whicli  is  impossible  unless  b  —  d  and  a  =  c. 

240.  If  X  and  y  are  real  and  x  +  y  V—  1  =  0,  then  x  =  0 
and  y  =  0. 

For,  2/ V—  l=  —  x, 

x'  +  f^O, 
which  is  true  only  when  x  =  0  and  y  =  0. 

241.  Operations  with  Imaginaries. 

(1)   Add5+7V^and8-9V"^i. 
The  sum  is  5  +  8  +  7  V^ -  9\/^l, 

or  13  -  2\/=^. 


(2) 

Multiply  3  +  2^ 

/^l  by  5  -  4V-  1. 

(3  +  2y/~l){b 

-4V-1) 

= 

15-12V-l  +  10V-l-8(- 

-1) 

= 

23-2V-1. 

(3)  Divide  14  4-5V 

-Iby2-3V-1. 

14^-5^/31 

(14  +  5\/iri)(2  +  3V-l) 

2_3V^ 

(2-3V~l)(2  +  3V:n) 

13  +  52\/-l 

4-(-9) 

13  +  52\/-  1 

13    . 

= 

1  +4V=:i. 

IMAGINARY    EXPRESSIONS. 


233 


Exercise  93. 
Reduce  to  the  form  5 V-  1  and  add  : 
1.   V^+V^=^.         6.   V^^  +  V- 4a"'  -  V-16 a*. 

3.   V-144+V-100.       8.  V-m  +  V^^  -  V^. 


4.  V- 256- V-16.      9.  3aV-4a''  +  2aV-49. 

5.  V^=T2l  -  V-^^^.    10.   Vl8  +  V-^^T8  -  V^. 

Reduce  to  the  form  bV-  1  and  multiply : 

11.  1  + V=^by  1- V^^. 

12.  4  +  V^  hy  4  -  V=^. 

13.  V3  -  2V^^  by  V3  +  2V^^. 

14.  V54  -  V^^  by  V54  +  V^=^. 

15.  V—  a  +  V—  6  by  V—  a  —  V—  6. 


16.  aV—  a'6*  by  aV—  a*b^. 

17.  2 V3  -  V^  by  2V3  +  V^5. 

18.  V^IO  by  V^^. 

Reduce  to  the  form  ^  V—  1  and  divide  : 

19.  V^^n[2by  V^.  23.  -  V25  by  V^^. 

20.  Vis  by  V="5.  24.  -  V^^^  by  -  V-^^. 

21.  V=^by  V-^=^.  25.  4V^'0  by  -  2 V^^^. 

22.  a  by  V^.  26.  4  +  V^  by  2  -  V^. 


CHAPTER    XIX. 

QUADRATIC   EQUATIONS. 

242.  We  have  already  considered  equations  of  the  first 
degree  in  one  or  more  unknowns.  We  now  proceed  to  the 
treatment  of  equations  containing  one  or  more  unknowns 
to  a  degree  not  exceeding  the  second.  An  equation  which 
contains  the  square  of  the  unknown,  but  no  higher  power, 
is  called  a  quadratic  equation. 

243.  A  quadratic  equation  which  involves  but  one  un- 
known number  can  contain  only : 

(1)  Terms  involving  the  square  of  the  unknown  number. 

(2)  Terms  involving  the  first  power  of  the  unknown 
number. 

(3)  Terms  which  do  not  involve  the  unknown  number. 

Collecting  similar  terms,  every  quadratic  equation  can 
be  made  to  assume  the  form 

ax^  -\-bx-{'C  —  0, 

where  a,  h,  and  c  are  known  numbers,  and  x  the  unknown 
number. 

If  a,  h,  c  are  numbers  expressed  by  figures,  the  equation 
is  a  numerical  quadratic.  If  a,  b,  c  are  numbers  represented 
wholly  or  in  part  by  letters,  the  equation  is  a  literal  quadratic. 

244.  In  the  equation  ax^  -\-hx-{-c  =  0,  a,  h,  and  c  are 
called  the  coefficients  of  the  equation.  The  third  term  c  is 
called  the  constant  term. 


QUADRATIC    EQUATIONS.  235 

If  the  first  power  of  x  is  wanting,  the  equation  is  a  pure 
quadratic ;  in  this  case  ^  =  0. 

If  the  first  power  of  x  is  present,  the  equation  is  an 
affected  or  complete  quadratic. 

Pure  Quadratic  Equations. 

245.    Examples. 

(1)  Solve  the  equation  bx^  —  48  =  2x\ 

We  have  bx^-4S  =  2x\ 

Collect   the  terms,  3x2  =  48. 

Divide  by  3,  a-^  =  16. 

Extract  the  square  root,  x  =  ±  4. 

It  will  be  observed  that  there  are  two  roots,  and  that  these  are 
numerically  equal,  but  of  opposite  signs.  There  can  be  only  two 
roots,  since  any  number  has  only  two  square  roots. 

It  may  seem  as  though  we  ought  to  write  the  sign  ±  before  the  x 
as  well  as  before  the  4.  If  we  do  this,  we  have  +  a;  =  +  4,  —  a;  =  —  4, 
+  a;  =  — 4,  -a;=+4. 

From  the  first  and  second  equations,  a;  =  4 ;  from  the  third  and 
fourth,  a;  =  —  4 ;  these  values  of  x  are  both  given  by  the  equation 
a:  =  +  4.  Hence  it  is  unnecessary  to  write  the  i  sign  on  both  sides  of 
the  reduced  equation. 

(2)  Solve  the  equation  Sx"^  —  15  =  0. 
We  have  3a;2=15, 

or  ic^  =  5. 

Extract  the  square  root,  x  =  ±  VS. 

The  roots  cannot  be  found  exactly,  since  the  square  root  of  5  can- 
not be  found  exactly ;  they  can,  however,  be  determined  approxi- 
mately to  any  required  degree  of  accuracy  ;  for  example,  the  positive 
square  root  of  5  lies  between  2.23606  and  2.23607. 

(3)  Solve  the  equation  3  a;'  +  15  =^  0. 
We  have  3x2  = -15, 

or  x*  =  —  5. 

Extract  the  square  root,  a;  =  ±  V—  5. 


236  ALGEBRA. 

There  is  no  square  root  of  a  negative  number,  since  the  square  of 
any  number,  positive  or  negative,  is  necessarily  positive. 

The  square  root  of  —  5  differs  from  the  square  root  of  +  5  in  that 
the  latter  can  be  found  as  accurately  as  we  please,  while  the  former 
cannot  be  found  at  all. 

246.  A  root  which  can  be  found  exactly  is  called  an 
exact  or  rational  root.  Such  roots  are  either  whole  numbers 
or  fractions. 

A  root  which  is  indicated  but  can  be  found  only  approx- 
imately is  called  a  surd  or  irrational  root.  Such  roots 
involve  the  roots  of  imperfect  powers. 

Rational  and  surd  roots  are  together  called  real  roots. 

A  root  which  is  indicated  but  cannot  be  found,  either 
exactly  or  approximately,  is  called  an  imaginary  root.    Such 
roots  involve  the  even  roots  of  negative  numbers. 
» 

Exercise  94. 

Solve : 

1.  .^^-3  =  46.  6.    5:^^-9-2:?;^  + 24. 

2.  2(:^^-l)-3(:r^+l)+14-0.    7.    (:r  +  2)^  =  4:r  +  5. 

3  6  2  '5  15  25   ■ 

4         3      ■       3     _Q  g     3a;' -27      90+4:r^_>^ 

'    l-\-x      l-x        '  '      :c'  +  3  x^  +  9 

rr       3         17  ,^     Q      ,  7      65a; 

5. -„  =  —  10.    8:^  +  -  — 

4a;'      6a;2      3  x        1 


11. 


12. 


4.r'  +  5      2a;'-5_7a.-^-25 
10  15  20 

10.^1;'+ 17      12a;'  +  2_5a;'-  4 
18  'lla;'-8'         9 


QUADRATIC   EQUATIONS.  237 

j3     14a;^+16       ^x'  +  S  _2x^ 
21  85;2-ll        3' 

14.    x^ -\-bx -{- a=hx{l  —  bx). 

15.    mx^-\-n  =  q.  16.    x^  —  ax -{-h  =  ax{x  ~\). 

Affected  Quadratic  Equations. 

247.  Since  {x  =b  hf  =  x^  ±2  bx  +  5^  it  is  evident  that 
the  expression  x^±:2bx  lacks  only  the  third  term,  b^,  of 
being  a  perfect  square. 

This  third  term  is  the  square  of  half  the  coefficient  of  x. 

Every  affected  quadratic  may  be  made  to  assume  the 
form  xi^  ±:2bx  =  c,  by  dividing  the  equation  through  by 
the  coefficient  of  a;^ 

To  solve  such  an  equation : 

The  first  step  is  to  add  to  both  members  the  square  of 
half  the  coefficient  of  x.    This  is  called  completing  the  square. 

The  second  step  is  to  extract  the  square  root  of  each 
member  of  the  resulting  equation. 

The  third  step  is  to  reduce  the  two  resulting  simple 
equations. 

(1)  Solve  the  equation  x^  ~Sx  —  20. 

We  have  x^-^x  =  20. 

Complete  the  square,      x^  —  8  x  +  16  =  36. 
Extract  the  square  root,  re  —  4  =  ±  6. 

Reduce,  re  =  4  +  6  =  10, 

or  a;  =  4-6  =  -2. 

The  roots  are  10  and  —  2. 

Verify  by  putting  these  numbers  for  x  in  the  given  equation  : 


a;  =  10, 

102-8(10)^20, 

100  -  80  =  20. 


a;  =  -2, 
(_  2)2 -8  (-2)  =  20, 
4  +  16  =  20. 


238 


ALGEBRA. 


(2)   Solve  the  equation 


x-1       x-^9' 


Free  from  fractions,  {x  +  1)  (x  +  9)  =  (a;  —  1)  (4  a;  —  3). 
Simplify,  3a;2-17x  =  6. 

We  can  reduce  the  equation  to  the  form  x^  —  2bxhy  dividing  by  3. 
Divide  by  3,  "-^  -  -^-^-  ^-'> 


X"  —  -V-  X  =  Z. 


Half  the  coeflScient  of  a;  is  |  of  —  y-  =  -  Y,  and  the  square  of  —  y 
is  -2/^^-.     Add  the  square  of  —  y  to  both  sides,  and  we  have 

3        V  t)  y  36 

2      17      ,   /17\'     361 

x^-y-+ (-)=-• 

Extract  the  square  root. 


)0t. 

X  - 

17_ 
"  6 

^f 

.    X  = 

17  ,19 
"  6  "^  6 

§§  =  6. 
6 

x  = 

17 
"  6  ' 

19 
6 

2_ 
6 

1 
"3 

The  roots  are  6  and 


Verify  by  putting  these  numbers  for  x  in  the  original  equation 


6  +  1      24-3 


6-1       6+9 

7^21 
5      15 


That  is, 


1  =  1 
5     5' 


X  = 

1 
3' 

1 
3 

+  1 

4 

3 

3 

1 
3 

-1 

9 

2_ 
4 

13 
26" 

That 

is. 

1 

2 

1 
2 

QUADRATIC   EQUATIONS.  239 

248.  When  the  coefficient  of  x"^  is  not  unity,  we  may- 
proceed  as  in  the  preceding  section,  or  we  may  complete 
the  square  by  another  method. 

Since  (ax  ±  by  is  identical  with  aV  ±  2abx-[-b'^,  it  is 
evident  that  the  expression  a'x^  ±2abx  lacks  only  the 
third  term,  b^,  of  being  a  perfect  square. 

This  third  term  is  the  square  of  the  quotient  obtained 
by  dividing  the  second  term  by  twice  the  square  root  of 
the  first  term. 

Every  affected  quadratic  may  be  made  to  assume  the 
form  aV  ±  2abx  =  c  (§  247). 

To  solve  such  an  equation  : 

The  first  step  is  to  complete  the  square ;  to  do  this,  we 
divide  the  second  term  by  twice  the  square  root  of  the  first 
term,  square  the  quotient,  and  add  the  result  to  both  mem- 
bers of  the  equation. 

The  second  step  is  to  extract  the  square  root  of  each 
member  of  the  resulting  equation. 

The  third  step  is  to  reduce  the  two  resulting  simple 
equations. 

249.  Numerical  Quadratics  are  solved  as  follows : 

(1)  Solve  the  equation  \Q>x^  ■\-bx  -^  =  1  x^  -  x  -\-  45. 

16x2  +  5a;  -  3  =  7a;2  -  .T  +  45. 
Simplify,  9.^2  + 6  a- =  48. 

Complete  the  square,  O.t^  +  6a;  +  1  =  49. 

Extract  the  square  root,  3  a;  +  1  =  ±  7. 

Reduce,  3a;  =  -l+7or-l-7; 

3a;  =  6  or -8. 
.-.  a;  =  2or-2g. 
Verify  by  substituting  2  for  x  in  the  equation 

16x2+ 5a;- 3  =  7a;' -a; +  45. 

16  (2)2  +  5  (2)  -  3  =  7  (2)2  -  (2)  +  45, 

64  +  10  -  3  =  28  -  2  +  45, 

71-71. 


240         ,  ALGEBRA. 

Verify  by  substituting  —  2|  for  x  in  the  equation 

1024_40_3^448  +  8^45, 
9         3  9        3 

1024  -  120  -  27  =  448  +  24  +  405, 

877  =  877. 

(2)  Solve  the  equation  3a;^  —  4ri:  =  32. 

Since  the  exact  root  of  3,  the  coefficient  of  x^,  cannot  be  found,  it 
is  necessary  to  multiply  or  divide  each  term  of  the  equation  by  3  to 
make  the  coefficient  of  x^  a  square  number. 

Multiply  by  3,  9  a;^  -  12  a;  =  96. 

Complete  the  square,        9a;'^  —  12  a;  +  4  =  100. 


Extract  the  square  root, 

3  a;  -  2  =  ±  10. 

Reduce, 

3«  =  2  +  10or2-10 

3a;  =  12  or -8. 

.-.  a:  =  4  or  -  2f . 

Or,  divide  by  3, 

"^        3        3 

Complete  the  square, 

,      4a;     4     32     4  _  100 
3       9      3      9       9 

Extract  the  square  root. 

3          3 

2  ±10 
•••"          3 

=  4or-2f. 

Verify  by  substituting  4  for  x  in  the  original  equation, 

48  -  16  =  32, 

32  =  32. 

Verify  by  substituting  -  2f  for  x  in  the  original  equation, 

21i  +  (10f)  =  32, 

32  =  32. 


QUADRATIC    EQUATIONS.  241 

(3)   Solve  the  equation  -  Sx'  -\-  5x  =  —  2. 

Since  the  even  root  of  a  negative  number  is  impossible,  it  is  neces- 
sary to  change  the  sign  of  each  term.     The  resulting  equation  is 

3a:2-5a;  =  2. 


Multiply  by  3,  9  .c^  _  15  a;  =  6, 

25^4; 
4       4 


Complete  the  square,  9x^  —  15x  -] 


5         7 
Extract  the  square  root,  3x  —  =  ±  -• 

Reduce,  3  a;  =     ~     ; 

3a;  =  6or-l. 
/.  re  =  2  or  —  -• 

■"  o 

Or,  divide  bv  3,  a:2-^==| 

3       3 

Complete  the  square,  x^ '-  +  Tr^'=  77,' 

3       36      36 

5         7 

Extract  the  square  root,  x ==  ±  -• 

^  6         6 

5±7 


2  or 


If  the  equation  3  a;^  —  5  a;  =  2  is  multiplied  by /our  times  the  coeffi- 
cient of  x^,  fractions  will  be  avoided  : 

36a^2_60a;  =  24. 
Complete  the  square,  36  x^  -  60  a:  +  25  =  49. 
Extract  the  square  root,  6  a;  —  5  =  ±  7. 

6a;  =  5±7. 

.-.  a;  =  2  or  —  -• 
o 

The  number  added  to  complete  the  square  by  this  last  method  is 
the  square  of  the  coefficient  of  x  in  the  original  equation  3a;^  —  5  a;  =  2. 

Note.  If  the  coefficient  of  x  is  an  even  number,  we  may  multiply 
by  the  coefficient  of  x^,  and  add  to  each  member  the  square  of  half  the 
coefficient  of  x  in  the  given  equation. 


242  ALGEBRA. 

o  1 

(4)  Solve  the  equation  —^ =  2. 

b  —  X     2x  —  b 

Simplify,  4  a;^  -  23  a;  =  -  30. 

Multiply  by  four  times  the  coefficient  of  x^,  and  add  to  each  side 
the  square  of  the  coefficient  of  x, 

64  :r2  -  ( )  +  (23)2  _  529  _  430  =  49. 
Extract  the  root,  8  a;  —  23  =  ±  7. 

Reduce,  8  a;  =  23  ±  7 ; 

8a; -30  or  16. 
.-.  X  =  3|  or  2. 

If  a  trinomial  is  a  perfect  square,  its  root  is  found  by  taking  the 
roots  of  the  first  and  third  terms  and  connecting  them  by  the  sign  of 
the  middle  term.  It  is  not  necessary,  therefore,  in  completing  the 
square,  to  write  the  middle  term,  but  its  place  may  be  indicated  as 
in  this  example. 

(5)  Solve  the  equation  72a;'  —  30:r  =  —  7. 

Since  72  =  2^  x  3^,  if  the  equation  is  multiplied  by  2,  the  coeffi- 
cient of  x"^  in  the  resulting  equation,  144 a;'^  —  60a;  =  —  14,  will  be  a 
square  number,  and  the  term  required  to  complete  the  square  will  be 
(60)2  ^  (5^2  ^  2^  Hence,  if  the  original  equation  is  multiplied  by 
4x2,  the  coefficient  of  x^  in  the  result  will  be  a  square  number,  and 
fractions  will  be  avoided  in  the  work. 

Multiply  the  given  equation  by  8, 

576x2 -240  a;  =  -56. 
Complete  the  square, 

576a;2-()  +  25  =  -31. 
Extract  the  root,  24  a;  —  5==+  V— 31. 

Reduce,  24a;  =  5±  V-31. 


.-.  a;  =  ,V(5±  V- 

-31). 

Solve : 

Exercise  95. 

1. 

x'-{-4x=12. 

4.    x'-7x==8. 

7.    rr2-a;  =  6. 

2. 

x'-6x=16. 

5.    3a;' -4a- -7. 

8.    5:r'-3:c=:2. 

3. 

x'-12x-}-Q  =  l 

-.  6.    12a;'+:r-l  =  0. 

9.    2:i;'-27:r=14. 

QUADRATIC    EQUATIONS.  243 

10.  x'-  —  +  ~  =  0.  13.    ^±l=^2a:-l 

3       12  x  +  4:       x+e> 

11.  ^-^^2(x  +  2).  14.       ^  07  +  3    _      1 


2      3         '  '  x-i-1      2(a;  +  4)         18 

12.    ^  +  A^1?.  15.    -2^--^+     2 


43a:       6  .'c-la;-2a;-4 

16.    bx(x-S)--2(x'-Q)==(x  +  ^)(ix  +  4:). 

17  3a;  5__    3a;^  23 

2(a;+l)   "8      a;^-l      4(a;-l)' 

18.  (.r-2)(a;-4)-2(a;-l)(a;-3)  =  0. 

19.  i(a;-4)--(a;-2)=-(2a;  +  3). 
7  5  a; 

20.  |(3a;^  -x-  ^)-hx'  -l)  =  2(x  -  2)1 

21     2a;  .    3a; -50   _  12a; +70 
15      3(10  +  a;)  190 


22. 


23. 


24. 


a; 

15 -7a; 
8(1-:.) 

25. 

14a;-9_ 

0.^-3 

a;'^-l 

8a;-3 

a;+l 

2a;-l  1   l_2a;- 
a;—  1       6       X  — 

f  - 

1       x  +  5 

2a;+l 

x-6 
x-2 

x-i-2 
x-1 

4-a;      7 
2a;        3 

27. 

X      ,  7  —  a; 
7  —  a;         X 

=  2A. 

^R 

2a;  +  3 

7 -a; 

7-3a; 

2(2a;-l) 

2(a:  +  l) 

4-3a; 

29. 
30. 

12a;^-lla; 
8a;^- 

3a; -1      5 

^+10a;- 
7a;  4- 6 

™=H.-| 

7-a;       2a;+l 


244  ALGEBRA. 

250.   Literal  Quadratics  are  solved  as  follows : 
(1)   Solve  the  equation  ax^  -^-hx^^c. 
Multiply  the  equation  by  4  a  and  add  the  square  of  6, 

4aV  +  0  +  62  ^4ac  +  5l 
Extract  the  square  root,  2aa;  +  6  =  ±  •\/4ac  +  6^ 

Reduce,  2aa:  =  —  &  +  \/4ac  +  W. 

—  b+\^iac-\-h^ 


(2)   Solve  the  equation  adx  —  acx"^  =  hex  —  hd. 
Transpose  hex  and  change  the  signs, 

acx^  +  hex  —  adx  =  hd. 
Express  the  left  member  in  two  terms, 
acx'^  +  {he  —  ad)  x  =  hd. 
Multiply  by  4ac, 

4  a'^e'^x'^  +  4  ac  {he  —  ad)  a;  =  4  ahed. 
Complete  the  square, 

4a'^cV  +  ( )  +  (6c  -  ad)""  =  h^c""  +  2ahcd  +  a'^cZ^. 
Extract  the  root,        2acx  +  (6c  —  ac?)  =  ±  (6c  +  acZ). 
Reduce,  2  acx  =  —  (6c  —  ad)  ±  {he  +  ad) 

=  2  ac?  or  —  26c. 

(^          6 
. .  iK  ==  -  or 


(3)   Solve  the  equation  px"^  —px-\-  qx^  -}-  qx  — 


p-hq 

Express  the  left  member  in  two  terms, 

PI 
{p  +  q)x^-{p-q)x=^^- 

Multiply  by  4  times  the  coefficient  of  x^, 

4:{p  +  qYx"^ —  4:{p'^  —  q'^)x  =  4pq. 
Complete  the  square, 

4{p  +  qf  .t2  _  ( )  +  (p  _  ^)2  =p2  +  2pq  +  q\ 
Extract  the  root,  2{p  +  q)  x  —  {p  —  q)  =  ±  (p  +  q). 
Reduce,  2{p  +  q)  x  =  {p  —  q)  ±  {p  +  q) 

=  2por  —  2q. 

p  q 

•    -  or — 


p+q      p+q 

Note.  The  left-hand  member  of  the  equation  when  simplified 
must  be  expressed  in  two  terms,  simple  or  compound,  one  term  con- 
taining x^,  and  the  other  term  containing  x. 


Solve  : 

1. 

x' 

-\-2ax 

=  a\ 

2. 

x' 

=  iax 

+  7a^ 

3. 

x' 

4 

—  3  7nx. 

4. 

x' 

bnx 

2 

2 

0, 

QUADRATIC   EQUATIONS.  246 

Exercise  96. 


14.  a^  -^  ax  =  a  -\-  X. 

15.  :r^  +  ao;  =  5a;  +  «^- 

16.  ^  +  «  =  ^4-^. 
a      X      b      X 

<,,.  1+1=1+1 


a;     a;  +  5      a     a-\-b 

18.  ^4-^-^  =  0. 

(rc  +  a)''      {x-ay  3       4       3a 

6.  c:u  =  a:rH5:i;'^--^.    19.  r^+-?  =  a  +  ^^- 

«  +  5  .T-3  .r  +  3 

7.  — — I —  = 20.  mx^  —  l=—^ ^• 

h'^       c^        c  TYin 

,8.    {c^  ^Y)x  =  ax^^a.         21.  {ax  —  l>){hx  —  a)  ^  & . 

„         a      ,       h     _    2c         oo<^^  +  ^_ "^^^  +  '^ 
y. (- — -•      Zai. — 

X  —  a     X  —  0      X  —  c  ox  -{-  a      nx-\-in 

10.    1 =  i  +  i  +  i-     23.  -i^+_i^  =  .. 

a-\-o  -{-x      a      0      X  m-{-  x     m  —  x 

11^    _1 l__A+i^.  24.  (^-l)V+2(3a-l>_Y 

a  —  a;      a  +  ^'      a^  —  a:''  4a  — 1 

^2_    x'+2a&(«'  +  y)^2x.      25.  (£5!^i!M+l}  =  2:.. 
a^  +  6"^  a'*  +  b"^ 

2x~a  +  2b  (m  +  nf        ^  ^ 

97       »  ,  a-6_14a'^-5a5-10^>'  ,  (2a-Sb)x 
^'    "^  +    a^^    "  18a^6^  "^        2a^> 


i^46 

ALGEBRA. 

28. 

ate^  +  *'^  = 

_6a'^  +  a6-252'     Za^x 

c 

c"                   c 

9Q 

x" 

ml~4:a^  _x 

3m-2a     Aa-6m     2 

30.  6^  +  ("  +  ^)'  =  5(a-&)  +  g^. 

a;  6:r 

31.  ^(x^  +  a''  +  ab)  =  lx(20a  +  ^b). 

32.  rc'^  —  (6  —  (x)  (?  =  a:?;  —  5:i:  +  ^^• 

33.  a;' —  2ma;  =  (w— jo  + 77i)(w  —  j9  —  m). 

34.  x"^  —  (in  -}-  n) X  =  \(p  -\-  q  -\-  m  -^  n)(p  -\-  q  —  m  —  n). 

35.  wma:^  —  (m  +  n){mn  -\- 1) x -\- {m -\- Tif  =  0. 

gg     2&  — rg  — 2a  ■  45  — 7a_a;  — 4a 
6:r  a:?;  —  6:r      ah  —  b'^ 

37.    2rr'^(a^  -  b'')  -  (3a^  +  ^>^)(:r  -  1)  =  (3 5=^  +  a'^)(a;  +  1). 

38     <^~^^~"^       5b  —  X    ■  2a  — a:— 195__^ 

a^  —  ^  b'^        ax-\-2bx          2bx  —  ax 

gg     a;+13a  +  3&      ■|^_a  — 25 
ba  —  ?>b  —  X  X  -\-2b 

4Q         a;  +  3^> 3^ a  +  35  _q 

8a^-12a5     9^)'^-4a^     (2 a  +  3 Z>)(a:  -  3 5)    '    ' 

41.  ?2:r^-{-i5^ — J^^'^  —  m,x-\-7n — 7i  =  Q. 

42.  (aH-^>  +  c)a;^-(2a  +  Z>  +  c)a:  +  a  =  0. 

43.  {ax  —  b){c  —  d)  —  (a  ~  b)(cx  —  d)x. 
^^  2a: +1      Ifl      2\_32:+l 


5 
2:rM-^-l  '  2x''-Zx-\-\      2bx-~b 


^g  1 L  -^  —  ^  2^57  +  5 


QUADRATIC    EQUATIONS.  247 

251.  Solutions  by  a  Formula.  Every  affected  quadratic 
may  be  reduced  to  the  form  x^ -{- px -}-  q  =  0,  in  which  ^ 
and  q  represent  numbers,  positive  or  negative,  integral  or 
fractional. 

Solve  :  x^  -j-px-\-  q  =  0. 

4:x'  +  ()+p'=p^-4:q, 

2x-\-p  =  ±  Vp^  —  4 5'. 

.'.  x  =  —  -zt  -  y/p'^  —  4:0. 
2      2 

By  this  formula,  the  values  of  x  in  an  equation  of  the 

form  x"^  -\-px  -}-  q  - -  0,  may  be  written  at  once.     Thus,  take 

the  equation 

3a;' -5a; +  2  =  0. 

Divide  by  3,  x^ a;  4-  -  =  0. 

^  3        3 

XT  5,2 

Here,  p  = --,  and  q  =  -• 

5 


± 
6     2 

5  1 

-  ±  - 

6  6 


1       2 
=  lor-. 


'7  \   9 


252.  Solutions  by  Factoring.  A  quadratic  which  has  been 
reduced  to  its  simplest  form,  and  has  all  its  terms  written 
on  one  side,  may  often  have  that  side  resolved  by  inspection 
into  factors. 

In  this  case  the  roots  are  seen  at  once  without  complet- 
ing the  square. 

(1)  Solve  a:' +  7  a; -60-0. 

Since  x^  +  7 x  -  60  =  (x  +  12){x-  5), 

the  equation  x^  +  7x  —  60  =  0 

may  be  written  (x  +  12)(x  —  5)  =  0. 


248  ALGEBKA. 

If  either  of  the  factors  a;  +  12  or  a;  -  5  is  0,  the  product  of  the  two 
factors  is  0,  and  the  equation  is  satisfied. 

Hence,  a;  +  12  =  0,  or  a;  -  5  =  0. 

.•.  x  =  —  12,  or  x  =  b, 

(2)  Solve  x'-\-lx  =  0. 

The  equation  a;^  +  7  x  =  0 

becomes  x{x  +  1)  =  0, 

and  is  satisfied  if  a;  =  0,  or  if  a;  +  7  =  0. 

.•,  the  roots  are  0  and  —  7. 

This  method  is  easily  applied  to  an  equation  all  the  terms  of  whicli 
contain  x. 

(3)  Solve  2a;' -a;^- 6a;  =  0. 

The  equation  2a;'  —  a;'^  —  6x  =  0 

becomes  x{2x'^  -  x  —  Q)  =  0, 

and  is  satisfied  if  x  =  0,  or  if  2  a;'^  —  a;  —  6  =  0. 

By  solving  2  x^  -  x  -  6  =  0  the  two  roots  2  and  -  -  are  found. 

2 

.•.  the  equation  has  three  roots,  0,  2,  —  -• 

(4)  Solve  x^-\-x'-^x-^  =  0. 

The  equation  a;'  +  x'^  —  4  a;  —  4  =  0 

becomes  x^{x  +  1)  —  4  (x  +  1)  =  0, 

(.x2-4)(x  +  l)  =  0.' 
.-.  the  roots  of  the  equation  are  —  1,  2,  —  2. 

(5)  Solve  .r' -  2a;' -  11  a;  +  12  =  0. 

We  find  by  trial  that  if  we  put  1  for  x,  the  equation  is  satisfied. 
Hence,  1  is  a  root. 

Divide  by  a;  —  1 ;  the  given  equation  may  be  written 
(a;-l)(a;2-a;-12)  =  0, 
and  is  satisfied         if  a;  -  1  =  0,  or  if  x^  -  a;  —  12  =  0. 

The  roots  are  found  to  be  1,  4,  —  3. 

(6)  Solve  the  equation  x{x''  —  9)  =  a(a'  -  9). 

If  we  put  a  for  x,  the  equation  is  satisfied ;  therefore  a  is  a  root 
(I  68). 


QUADRATIC   EQUATIONS.  249 

Transpose  all  the  terms  to  the  left  member,  and  divide  by  a;  —  a. 
The  given  equation  may  be  written 

and  is  satisfied  if  a;  —  a  =  0,  or  if  a'^  +  ax  +  a^  —  9  =  0. 
The  roots  are  found  to  be 


a,  ^(-a+ V36-3a2),  i  (_  a- Vse-Sa'-*). 

Exercise  97. 
Find  all  the  roots  of: 

1.  (x  +  lXa:-~2Xx'  +  x  —  2)  =  0.  7.  x^-x'-x-{-l  =  0. 

2.  (x'-Sx  +  2Xx'-x-12)  =  0.  8.  8a;'-l  =  0. 

3.  (x-\-lXx-2Xx  +  S)  =  -6.  9.  8a;'+l  =  0. 

4.  2a;' +  4a;'- 70a;  =  0.  10.  .7;«-l=0. 

5.  (a;' -  a;  -  6)(a;' -  a;  -  20)  =  0.  11.  x(x-aXx'-b')=0. 

6.  a;(a;+l)(a;-[-2)  =  a(a4-l)(a+2).  12.  7i(x'-\-l)-\-x-\-1^0. 

253.  Equations  in  the  Quadratic  Form.  An  equation  is  in 
the  quadratic  form  if  it  contains  but  two  powers  of  the 
unknown  number,  and  the  exponent  of  one  power  is  exactly 
twice  that  of  the  other  power. 

254.  Equations  not  of  the  second  degree,  but  of  the 
quadratic  form,  may  be  solved  by  completing  the  square. 

(1)    Solve:  8 x'  + 63x^=^8. 

We  have  8a;«  +  63a;3  =  8. 

Multiply  by  32  and  complete  the  square, 

256a;6  +  ()  + (63)2  =  4225. 
Extract  the  square  root,    16  a;'  +  63  =  i  65. 

Hence,  x*  =  -  or  —  8. 

8 


250 


ALGEBRA. 


Extracting  the  cube  root,  two  values  of  x  are  ^  and  —  2.     To  find 
the  remaining  roots,  it  remains  to  solve  completely  the  two  equations 
x3  =  |,        ■     a;3=-8. 

We  have,  8a^-l-0, 

or,      (2a'-l)(4a;' +  2x4-1) -0. 

.-.  either  2a;- 1  =  0, 

or,  4x2  + 2a; +  1=0. 

Solving  these,  we  find  for  three 
values  of  x. 


h 


]  +  V-  3     _  1  -  V-  3 


We  have,  a;^  +  8  =  0, 

or,  (a;  +  2)(a;2_2a;  +  4)  =  0. 

,-.  either  a;  +  2  =  0, 

or,  x^- 2a; +  4  =  0. 

Solving  these,  we  find  for  three 
values  of  x, 


2,     1  +  V-  3.     1 


3. 


4  4 

These  six  values  of  x  are  the  six  roots  of  the  given  equation. 

(2)  Solve:  V^'-- 3a/?  =  40. 

Using  fractional  exponents,  we  have  x^  —  3  x^  =  40. 
Complete  the  square,     4x^  -  12x''  +  9  =  169. 
Extract  the  square  root,  2x^  —  3  =  ±13. 

.-.  2x^  =  16,  or  -  10. 

a;4  =  8,  or  -  5, 
x=  16,  or  —by/b. 
There  are  other  values  of  x  which  we  do  not  at  present  find. 

(3)  Solve:  x" -{---\- x -\--=^  4:. 


Add  2  to  both  members, 


2  f  4.  +  a;  + 


Put  X  +  -  =  y  ;  the  equation  becomes 

X 

2/2  +  y  =  6. 

Solving  this,  y  =  2,  or  —  3. 

.-.  X  +  -  =  2,  or  X  +  -  =  -  3. 

X  X 

Solving  these  two  equations,  we  find  for  the  four  values  of  x, 


1.    1. 


3  +  V5     -3-\/5 


QUADRATIC    EQUATIONS.  251 

(4)  Solve:         x'  -  4:x'  -{-  5x'  -2x  -20  =  0. 
Begin  by  attempting  to  extract  the  square  root. 
X* -  ict^  -{-  5x^  -  2x  -  20\x^ -  2x 


X' 


2x2  _  2a; 


-4a;' +  5x2 
-ia^  +  4:X^ 


x^-2x-20. 
Hence,  the  equation  may  be  written  in  the  quadratic  form 

(a;2-2a:)2  +  a;2-2x-20  =  0. 
Put  x^  —  2x  =  y  \  the  equation  becomes 

2/2  +  2/  -  20  =  0. 
Solving  this,  y  =  —  5,  or  +  4. 

.•.  a;2  —  2a;  =  —  5,  or  a;2  —  2aj  =  4. 
Solving  these  two  equations,  we  find  for  the  four  values  of  x, 
l  +  2\/^,    1-2V^,    1  +  V5,   1-V5. 

Exercise  98. 

Solve  : 

1.  a;«  +  7a:'  =  8.  8.  (a;^-9)»=:3fll  (x^- 2). 

2.  :r*-5a;^  +  4-0.  9.  a;«  +  14a;'  + 24  =  0. 

3.  37a;'  — 9  =  4:r*.  10.  19a;*  + 216a;^  =  a;. 

4.  16a;«=17a;*-l.  11.  :r'  + 22a;*  + 21 -=  0. 

5.  32a;'''  — 33a;*4-l-=0.  12.  0;'"*  + Sa;**  — 4  =  0. 

6.  (a;'-2)'  =  i(a;'  +  12).  13.  4a;*-20a;«  +  23a;'+5a:=6. 

7.  ^.4„_5^__25_Q  ^^^    _L4_A_20  =  0. 

3         12  x'"     a;" 

15.  a;*~4r'-10a;'  +  28a;-15  =  0. 

16.  a;*-2a;'-13a;2+14a;  =  -24. 

17.  108a:*  =  20a;(9a;'-l)-51a;'  +  7. 

18.  (a;'  -  l)(a;'  -  2)  +  {x"  -  ^){x''  -  4)  -  x'  +  5. 


252  ALGEBRA. 

255.   Kadical  Equations,     If  an  equation  involves  radical 
expressions,  we  may  first  clear  of  radicals  as  follows : 


Solve  V^+4+ V2a;4-6=  V7a7  +  14. 
Square  both  sides, 


a;  +  4  +  2V{x  +  i){2x  +  6)  +  2a;  +  6  =  7a;  +  14. 
Transpose  and  combine,       2V{x  +  4)(2a;  +  6)  =  4a;  +  4. 
Divide  by  2  and  square,  (x  +  4)(2a;  +  6)  =  (2x  +  2)'^ 

Multiply  and  reduce,  x'^  —  3  a;  =  10. 

Hence,  x  =  5,  or  —  2. 

Of  these  two  values,  only  5  will  satisfy  the  original  equation. 
The  value  —  2  will  satisfy  the  equation 

Va;  +  4  —  V'2  a;  +  6  =  VT  a;  +  14. 

In  fact,  squaring  both  members  of  the  original  equation  is  equiva- 
lent to  transposing  V7x  +  14  to  the  left  member,  and  then  multiply- 
ing by  the  rationalizing  factor  Va;  +  4  -|-  V'2a;  -1-6-1-  y/lx  +  14,  so 
that  the  equation  stands 

{Vx  +  4:  +  V2x+ii-V7x  +  14)(Va;  -h  4  -f  V2a;  +  6  -I-  V7a;  -f  14)  =  0, 

which  reduces  to  ■\/(a;  +  4)(2a;  -f  6)  —  (2  a;  -f  2)  =  0. 

Transposing  and  squaring  again  is  .equivalent  to  multiplying  by 


(VxTi  -  V2x  +  6  +  V7x  +  U){Vx  +  4  -  V2a;-h6  -  V7x  +  14). 
Multiplying  and  reducing,  we  have 

a;2-3a;-10  =  0. 
Therefore,  the  equation  a;'^  —  3  a;  —  10  =  0  is  really  obtained  from 


(Va;  4-4  +  ■yJ2x  -f  6  —  V? a;  -f  14) 


X  (Va;  -h  4  +  V2a;  +  6  +  V7a;  +  14) 


X  (Va;  -f  4  -  V2a;  -f  6  -  V7a;  4-  14) 

X  ( Vx+l -  V2a;-|-6  -1-  V7a;-f  14)  =  0. 

This  equation  is  satisfied  by  any  value  that  will  satisfy  any  one 
of  the  jour  factors  of  its  left  member.  The  first  factor  is  satisfied 
by  5,  and  the  last  factor  by  —  2,  while  no  values  can  be  found  to 
satisfy  the  second  or  third  factor. 


QUADRATIC    EQUATIONS.  253 

Hence,  if  z  radical  equation  of  this  form  is  proposed  for  solution, 
if  there  is  a  value  of  x  that  will  satisfy  the  particular  equation  given, 
that  value  must  be  retained,  and  any  value  that  does  not  satisfy  the 
equation  given  must  be  rejected.  (See  Wentworth,  McLellan,  and 
Glashan's  Algebraic  Analysis,  pp.  278-281.) 

256.    Some  radical  equations  may  be  solved  as  follows : 

Solve  1x'-bx-\-8^7x'-bx+l  =  ~8. 

Add  1  to  both  sides, 

7a;2  -  5a;  +  1  +  8-\/7x2_5a;+  1  =  -  7. 


Put  y/lx^  —  5  a;  +  1  =  2/ ;  the  equation  becomes 

2/2 +  82/ =  -7. 
Hence,  y  =  —  l,or  —  7, 

We  now  have  7a;2  —  5a;  +  1  =  1,  or  7x2  —  5a;  +  1  =  49. 
Solving  these,  we  find  for  the  values  of  x, 

7  1'      7 
These  values  all  satisfy  the  given  equation  when  we   take   the 
negative  value  of  the  square  root  of  the  expression    7a;2  — 5a;  +  l; 
they  are  in  fact  the  four  roots  of  the  biquadratic  obtained  by  clear 
ing  the  given  equation  of  radicals. 


Solve 


Exercise  99. 

1.  x""  -3x-  eVx'  -  3a;  -  3  4-  2  =  0. 

2.  x' i-^^ --+-,-=  ^■ 

X     x^      36 

3.  {2x'-?>xY-2{2x^-2>x)  =  lb. 

4.  {ax  —  6)2  -f  4  a  {ax  —- b)  =  — — 

5.  3 {2x'  -  x)  -  {2x'  -  x)^  -  2. 

6.  Wx~Sx'-{-Hx'-5x  +  5)^  =  16. 

7.  x' i- X-' -{- X  i- X-' =  i.        9.    x'+x-{-i{x'  +  x)^  =  l 

8.  x'  +  V^ff^  =  19.  10.    {x^-i- 1)^  -i-ix-  1)^  =  5. 


254  ALGEBRA. 

11.    ^--1  =  2  +  25;"^.         12.    V3:^+5-  V3a;-5  =  4. 

13.  (x'-}-l)-x(x'-{-l)  =  -2x\ 

14.  2x'  -  2V2:r*^  -bx  =  5(x-\-  3). 

15.  :r  +  2-4a,V^T2=-12xl 

16.  V2:i;  +  a  +  V2a;  —  a  =  5. 

17.  V9x'  +  21a;+  1  -  V9x'  +  6:^;  +  1  =  3a;. 

18.  :r3  —  4a;^  +  a;"^  +  4  :r~^  =-  -  J. 

19.  V^r+l  +  V.X-  +  16  =  ^x  -f  25. 

20.  V2:i,-  +  1  —  Vo;  +  4  =  i  V:r  -  3. 

21.  Vo;  +  3  +  Va;  +  8  =  5  V^. 

22.    V3+^+V^  =  — 23.    Vx'-li-6  =  - 


16 


V3  +  a;  V:r^--1 


24.    — i_+        1  2 


V:?:  -f-  1       'Vx  —  1       -Vx'^  —  1 

/  25     V^  +  2  g  —  V:^^  —  2a  _  x 

Va;-2a+ V:r  +  2a      2a 

2g^    3a;+V4:r-3;^_o       ^^^    V7:r'H4  +  2V3:g- 1  _  ^ 
3a:-V4a;-a:''        *  '    V7^?+4-2V3:r- 1 

28.  V(a;  -  a)''  4-  2a6  +  Z^=^  =  a;  -  a  +  ^>. 

29.  V(a;  +  a)"'^  +  2a6  +  6"'  =  Z>  —  a  —  a;. 

31.    4a;^  -  3(a;^  +  l)(a;^  -  2)  =  a:^(10  -  3a;^). 
5.    {x^  -  2){x^  -  4)  -  a;^  (x^  -  1)^  -  12. 


32. 


QUADRATIC   EQUATIONS.  255 

257.  Problems  involving  Quadratics.  Problems  which  in- 
volve quadratic  equations  apparently  have  two  solutions, 
since  a  quadratic  equation  has  two  roots.  When  both  roots 
of  the  quadratic  equation  are  positive  integers,  they  will 
give  two  actual  solutions  of  the  problem. 

Fractional  and  negative  roots  will  in  some  problems  give 
admissible  solutions ;  in  other  problems  they  will  not  give 
admissible  solutions. 

No  difficulty  will  be  found  in  selecting  the  result  which 
belongs  to  the  particular  problem  we  are  solving.  Some- 
times, by  a  change  in  the  statement  of  the  problem,  we  may 
form  a  new  problem  which  corresponds  to  the  result  that 
was  inapplicable  to  the  original  problem. 

Imaginary  roots  indicate  that  the  problem  is  impossible. 

Here  as  in  simple  equations  x  stands  for  an  unknown 
number. 

(1)  The  sum  of  the  squares  of  two  consecutive  numbers 
is  481.     Find  the  numbers. 

Let  X  =  one  number, 

and  a;  +  1  =  the  other. 

Then  a;2  ^  (x  +  l)^  =  481, 

or  2a;2  +  2a;  +  1  =  481. 

The  solution  of  which  gives  a;  =  15  or  —  16. 

The  positive  15  gives  for  the  numbers,  15  and  16. 

The  negative  root  — 16  is  inapplicable  to  the  problem,  as  consecu- 
tive numbers  are  understood  to  be  integers  which  follow  one  another 
in  the  common  scale  1,  2,  3,  4 

(2)  A  pedler  bought  a  number  of  knives  for  $2.40. 
Had  he  bought  4  more  for  the  same  money,  he  would  have 
paid  3  cents  less  for  each.  How  many  knives  did  he  buy, 
and  what  did  he  pay  for  each  ? 

Let                                X  =  number  of  knives  he  bought. 
Then  =  number  of  cents  he  paid  for  each. 

X 


240 

X 

7  =  the  difference  in  price. 

X  +  4: 

3  =  the  difference  in  price. 

240_ 

240       o 

a; 

a;  +  4 

a;  =  16  or  -  20. 

256  ALGEBRA. 


But  if  a;  +  4  =  number  of  knives  he  bought, 

=  number  of  cents  he  paid  for  each, 

a;  +  4  ^ 


But 


Solving, 

He  bought  16  knives,  therefore,  and  paid  -^\^-,  or  15 
cents  for  each. 

If  the  problem  is  changed  so  as  to  read :  A  pedler  bought 
a  number  of  knives  for  $2.40,  and  if  he  had  bought  4  less 
for  the  same  money,  he  would  have  paid  3  cents  more  for 
each,  the  equation  will  be 

240      240  ^  3 
a;  —  4        X 
Solving,  a;  =  20  or  -  16. 

This  second  problem  is  therefore  the  one  which  the  neg- 
ative answer  of  the  first  problem  suggests. 

(3)  What  is  the  price  of  eggs  per  dozen  when  2  more  in 
a  shilling's  worth  lowers  the  price  1  penny  per  dozen  ? 

Let  X  =  number  of  eggs  for  a  shilling. 

Then  -  ==  cost  of  one  egg  in  shillings. 

X 

12 

and  —  =  cost  of  1  dozen  in  shillings. 

X 

But  if  a;  +  2  =  number  of  eggs  for  a  shilling, 

12 
=  cost  of  1  dozen  in  shillings. 

X  -\-  A 

12        12         1 
.-. —  =  —  (1  penny  being  j\  of  a  shilling). 

X         X  -T  u        iZ 

The  solution  of  which  gives  x  =  16,  or  —  18. 

And  if  16  eggs  cost  a  shilling,  1  dozen  will  cost  9  pence. 

Therefore,  the  price  of  the  eggs  is  9  pence  per  dozen. 


QUADRATIC   EQUATIONS.  257 

Exercise  100. 

1.  The  sum  of  the  squares  of  three  consecutive  numbers 

is  365.     Find  the  numbers. 

2.  Three  times  the  product  of  two  consecutive  numbers  ex- 

ceeds four  times  their  sum  by  8.    Find  the  numbers. 

3.  The  product  of  three  consecutive  numbers  is  equal  to 

three  times  the  middle  number.    Find  the  numbers. 

4.  A  boy  bought  a  number  of  apples  for  16  cents.     Had 

he  bought  4  more  for  the  same  money  he  would 
have  paid  i  of  a  cent  less  for  each  apple.  How 
many  did  he  buy  ? 

5.  For  building  108  rods  of  stone-wall,  6  days  less  would 

have  been  required  if  3  rods  more  a  day  had  been 
built.     How  many  rods  a  day  were  built  ? 

6.  A  merchant  bought  some  pieces  of  silk  for  $900.  -Had 

he  bought  3  pieces  more  for  the  same  money  he 
would  have  paid  $15  less  for  each  piece.  How 
many  did  he  buy  ? 

7.  A  merchant  bought  some  pieces  of  cloth  for  $168.75. 

He  sold  the  cloth  for  $12  a  piece  and  gained  as 
much  as  1  piece  cost  him.  How  much  did  he  pay 
for  each  piece  ?  • 

8.  Find  the  price  of  eggs  per  score  when  10  more  in  62J 

cents'  worth  lowers  the  price  31 J  cents  per  hundred. 

9.  The  area  of  a  square  may  be  doubled  by  increasing  its 

length  by  6  inches  and  its  breadth  by  4  inches.  De- 
termine its  side. 

10.  The  length  of  a  rectangular  field  exceeds  the  breadth 
by  1  yard,  and  the  area  is  3  acres.  Determine  its 
dimensions. 


258  ALGEBRA. 

11.  There  are  tliree  lines  of  which  two  are  each  ^  of  the 

third,  and  the  sum  of  the  squares  described  on  them 
is  equal  to  a  square  yard.  Determine  the  lengths 
of  the  lines  in  inches. 

12.  A  grass  plot  9  yards  long  and  6  yards  broad  has  a 

path  round  it.  The  area  of  the  path  is  equal  to 
that  of  the  plot.     Determine  the  width  of  the  path. 

13.  Find  the  radius  of  a  circle  the  area  of  which  would  be 

doubled  by  increasing  its  radius  by  1  inch. 

14.  Divide  a  line  20  inches  long  into  two  parts  so  that  the 

rectangle  contained  by  the  whole  and  one  part  may 
be  equal  to  the  square  on  the  other  part. 

15.  A  can  do  some  work  in  9  hours  less  time  than  B  can 

do  it,  and  together  they  can  do  it  in  20  hours. 
How  long  will  it  take  each  alone  to  do  it  ? 

16.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both  to- 
gether in  2  hours  55  minutes.  How  long  will  it 
take  each  pipe  alone  to  fill  the  vessel  ? 

17.  A  vessel  which  has  two  pipes  can  be  filled  in  2  hours 

less  time  by  one  than  by  the  other,  and  by  both  to- 
gether in  1  hour  52  minutes  30  seconds.  How  long 
will  it  take  each  pipe  a]pne  to  fill  the  vessel? 

18.  An  iron  bar  weighs  36  pounds.     If  it  had  been  1  foot 

longer  each  foot  would  have  weighed  ^  a  pound  less. 
Find  the  length  and  the  weight  per  foot. 

19.  A  number  is  expressed  by  two  digits,  the  units'  digit 

being  the  square  of  the  other,  and  when  54  is  added 
its  digits  are  interchanged.     Find  the  number. 

20.  Divide  35  into  two  parts  so  that  the  sum  of  the  two 

fractions  formed  by  dividing  each  part  by  the  other 
may  be  2^j. 


QUADRATIC   EQUATIONS.  259 

21.  A  boat's  crew  row  3^  miles  down  a  river  and  back 

again  in  1  hour  40  minutes.  If  the  current  of  the 
river  is  2  miles  per  hour,  determine  their  rate  of 
rowing  in  still  water. 

22.  A  detachment  from  an  army  was  marching  in  regular 

column  with  5  men  more  in  depth  than  in  front. 
On  approaching  the  enemy  the  front  was  increased 
by  845  men,  and  the  whole  was  thus  drawn  up  in 
5  lines.     Find  the  number  of  men. 

23.  A  jockey  sold  a  horse  for  $144,  and  gained  as  much 

per  cent  as  the  horse  cost.    What  did  the  horse  cost? 

24.  A  merchant  expended  a  certain  sum  of  money  in  goods, 

which  he  sold  again  for  $24,  and  lost  as  much  per 
cent  as  the  goods  cost  him.  How  much  did  he  pay 
for  the  goods  ? 

25.  A  broker  bought  a  number  of  bank  shares  ($  100  each), 

when  they  were  at  a  certain  per  cent  discount^  for 
$7500  ;  and  afterwards  when  they  were  at  the  same 
per  cent  premium,  sold  all  but  60  for  $  5000.  How 
many  shares  did  he  buy,  and  at  what  price  ? 

26.  The  thickness  of  a  rectangular  solid  is  f  of  its  width, 

and  its  length  is  equal  to  the  sum  of  its  width  and 
thickness;  also,  the  number  of  cubic  yards  in  its 
volume  added  to  the  number  of  linear  yards  in  its 
edges  is  f  of  the  number  of  square  yards  in  its  sur- 
face.    Determine  its  dimensions. 

27.  If  a  carriage-wheel  16J  feet  round  took  1  second  more 

to  revolve,  the  rate  of  the  carriage  per  hour  would  be 
1|  miles  less.  At  what  rate  is  the  carriage  travelling? 


CHAPTER  XX. 
SIMULTANEOUS  QUADRATIC   EQUATIONS. 

258.  Quadratic  equations  involvi'ng  two  unknown  num- 
bers require  different  methods  for  their  solution,  according 
to  the  form  of  the  equations. 

Case  I. 

259.  When  from  One  of  the  Equations  the  Value  of  One  of 
the  Unknown  Numbers  can  be  found  in  Terms  of  the  Other,  and 
this  Value  substituted  in  the  Other  Equation. 

Solve:  3a;^-2a:y  =  5)  (1) 

x-y^^S  (2) 

Transpose  x  in  (2),  y  =  a;  —  2. 

In  (1)  put  a;  —  2  for  3/. 

3a;2-2x(x-2)  =  5. 
The  solution  of  which  gives  a;  =  1,  or  a;  =  —  5, 

If  0^  =  1, 

2/=l-2  =  -l; 
and  if  a;  =  —  5, 

2/  =  -5-2  =  -7. 
We  have  therefore  the  "pair^  of  values, 

The  original  equations  are  both  satisfied  by  either  pair  of  values. 
But  the  values  a?  =  1,  y  =  —  7,  will  not  satisfy  the  equations ;  nor  will 
the  values  a;  =  —  5,  y  =  —  1. 

The  student  must  be  careful  to  join  to  each  value  of  x 
the  corresponding  value  of  y. 


SIMULTANEOUS   QUADRATIC   EQUATIONS. 


261 


Case  II. 

260.  When  the  Left  Side  of  Each  of  the  Two  Equations  is 
Homogeneous  and  of  the  Second  Degree. 


Solve  : 


a;^  =  16  3 


(1) 

(2) 


17 

2i;2- 

-4v  +  3 

16. 

16 

v2_ 

1 

16 

Let  y  =  vx,  and  substitute  vx  for  y  in  both  equations. 
From  (1),  2«V  _^^^^^Zx^  =  17. 

From  (2),  v^a^^  - 

Equate  the  values  of  a:^,  ^^,_^^^  3     ^2  _  1 

32  v2_  64^  +  48 -17^2 -17, 
15v2_64v  =  -65, 
225^2 -960v  =  - 975,      y 
225^2-0 +  (32)2  =  49, 
15v-32  =  ±7. 

5       13 

•••^  =  3'^^T 

If 


If. 

1;  = 

5 

3/  = 

=  v*  = 

5a; 
^  3' 

Substitute  in 

(2). 

9 

-a;2  = 

=  16, 

x2  = 

=  9, 

iC  = 

=  ±3, 

y  = 

5a; 
"  3  ' 

=  ±5. 

13 
13a; 

y  =  VX  =  —- — 

^  5 


Substitute  in  (2), 
169x2  _ 
25 


16. 


2     25 


x  =  ± 
13  a; 


5 
3' 
13 


y  =  — ="3 


262  ALGEBRA. 


Case  III. 


261.   When  the  Two  Equations  are  Symmetrical  with  Respect 
to  X  and  y ;  that  is,  when  x  and  y  are  similarly  involved. 

Thus,  the  expressions 

are  symmetrical  expressions.  In  this  case  the  general  rule 
is  to  combine  the  equations  in  such  a  manner  as  to  remove 
the  highest  powers  of  x  and  y. 

Solve:  a;*  +  y*  =  337)  (1) 

a:+3/=      7j  (2) 

To  remove  x^  and  3/*,  raise  (2)  to  the  fourth  power, 

a;4  +  4  a;3y  +  6  xY  +  4  0^2/3  +    y^  =  2401 
Add(l),         a;*  +    y' =   337 

2  a;*  +  4  «3y  +  6  xY  +  4:xy^  +  2y^  =  2738 
Divide  by  2,  a*  +  2  x^y  +  3  xY  +  2xy^+y*  =  1369. 
Extract  the  square  root,  x"^  +  xy  +  y^  =  ±  37.  (3) 

Subtract  (3)  from  (2)2,  xy  =-- 12,  or  86. 

We  now  have  to  solve  the  two  pairs  of  equations, 


From  the  first, 


From  the  second, 


x  +  y=    71      x  +  y  = 

'  n 

xy  =  12)'        xy  =  SQi 

-  =  t];      or      ^  =  ^|. 

^      7  ±  V-  295 
2 

7  T  V-  295 

y         9 

SIMULTANEOUS    QUADRATIC   EQUATIONS. 


263 


262,  The  preceding  cases  are  general  methods  for  the 
solution  of  equations  which  belong  to  the  kinds  referred  to ; 
often,  however,  in  the  solution  of  these  and  other  kinds  of 
simultaneous  equations  involving  quadratics,  a  little  inge- 
nuity will  suggest  some  step  by  which  the  roots  may  be 
found  more  easily  than  by  the  general  method. 


tf 


itli-^- 


(1)  Solve:  ^  +  y=   40 j 

a;y  =  300  3 

Square  (1),  x^  +  2xy  +  y"^  ^  1600. 

Multiply  (2)  by  4,  4.xy  =  1200 

Subtract  (4)  from  (3), 

a;2- 2x2/ +  2/2  =  400. 
Extract  root  of  each  side,       a;  —  y  =  ±  20. 


IL 


L-!- 


From  (1)  and  (6), 

X  =  30  ) 
2/  =  10  i  ' 

(2)  Solve  : 

X  ^  y       20 

1,1       41 

:x?'^  f     400  J 

Square  (1), 

1^2      1^81. 
x^     xy     2/2     400 

Subtract  (2)  from  (3), 

2       40 
xy     400 

Subtract  (4)  from  (2), 

1       2       1  _    1 
x^     xy     f     400 

Extract  the  root, 

a;     2/        20 

From  (1)  and  (5), 

.  p}- 

^>^ 


a;=10-| 


y 


30  / 


(1) 
(2) 

(3) 
(4) 

(5) 
(6) 


(1) 
(2) 

(3) 
(4) 


(5) 


x  =  h\ 
3/  =  4i 


264  ALGEBRA. 


(3)  Solve  :  x  —y  =    4:) 

^2  +  y^  =  40| 

Square  (1),  x^  -2xy  +  y^  =  16. 

Subtract  (2)  from  (3),        -  2a;y  =  -  24. 
Subtract  (4)  from  (2), 

x^  +  2xy  +  2/^  =  64. 
Extract  the  root,  a;  +  y  =  ±  8. 

?rom  (1)  and  (5),  a;  =  61  x  =  -2 

y  =  2i'  ""'  y  =  -<d 


(4)  Solve :                 rr^  +  y^  =  91 1  (1) 

x--\-y  =    7j  ^2) 

Divide  (1)  by  (2),      x^-xy  +  ?/2  =  13.  (3) 

Square  (2),              a;^  +  2  a;?/ +  2/^  =  49.  '                   (4) 

Subtract  (3)  from  (4),  Zxy  =  36. 

Divide  by  -  3,                      -.xy  =  - 12.  (5) 

Add  (5)  and  (3),     x^  ~  2xy  +  y"^  =1. 

Extract  the  root,                 x  —  y  =  ±\.  (6) 

From  (2)  and  (6),  a;  =  41^^aj  =  3| 

(5)  Solve:                 x^'^-y^^^l^xy')  (1) 

X  -{-y  -12   ^  j                                (2) 

Divide  (1)  by  (2),     x^  -  xy  +  y'^  =  ^.  (3) 

Square  (2),               a;^  +  2  xy  +  y^  =--  1 14.  (4) 


Subtract  (4)  from  (3), 

-3.y  =  ^-144, 

lich  gives 

xy  =  2>2. 

We  now  have, 

x^-y  =  l2\ 
xy  =  ^2f 

Solving,  we  find, 

2/  =  4/'        y=8 

SIMULTANEOUS    QUADRATIC   EQUATIONS. 


265 


Exercise  101. 

Solve  : 

1.    x  +  y  =  lS') 
xy  =  SQ       i 

11. 

a;  -f-  y  =  49        1 
a;2  +  y^  =  168lj 

2.    x-{-y  =  29) 
xy  =  100     j 

12. 

x'  +  f  =  34:11 
xi-y=U      j 

3.   x-y  =  l9) 
xy==6Q       j 

13. 

a;  +  y=12        j 

4.    x  —  y^4:b 
xy=250      ' 

14. 

r^-/-98) 
x-y=2      i 

5.   x-y----10      ) 

15. 

x'  -  f  =  279  I 
a;-y-3       j 

6.    x-y-^U      I 
x^  +  f  =  436  j 

16. 

:r-3y  =  l| 
xy  +  f=b] 

7.   x  +  y  =  12     ■) 
x'-\-y'=104:\ 

17. 

4y=-5x+l    ) 
2a:y-33~:i'M 

8.  !+!=§    1 

X      y      i 
x'^y'     16  J 

18. 

1-1^3      ^ 

X     y 

9.    1  +  1=.5      ^ 

X     y 

19. 

1-1  =  2^    1 

X     y 

i-i  =  8| 

10.    7x'^-8xy  = 
5a;  +  2y  =  7 

159 

! 

20. 

o;'  —  2a;y  —  y 
x-\-y  =  2 

^=1 

} 

266  ALGEBRA. 

Exercise  102. 
Solve  : 

1.  x^^-xy-^-ly'^n^    )  9.    2a;' +  3a:y +  y^  =  70  ) 
2:r'4-2a;y  +  2/^=73  j  6 :?;' +  a;y  -  y' =  60    j 

2.  :r'4-a:y  +  4y'  =  6)  10.    x'-xy-y'^h         ) 
3a;'  +  8y'=14       j  2a;'  + 3a:y +  y'  =  28  J 

3.  ;r'-:r?/  +  y^  =  21)  11.    ^xy  =  ^^-xY\ 
y''-%xy  =  -\fi  j  :r  +  y=:6             J 

4.  :r'-4y'^-9-0|  12.    :^'^  +  y' =  18  -  a; -y  ) 
a;y  +  2y'-3=0J  xy.=--^                         j 

5.  a:'^~-^y-35  =  0|  13.    2  (^r^  +  y')  =  5  a;y  ) 
a:y  +  y^-18  =  0j  4(a;-y)=a;y       j 

6.  .T'  +  a:y  +  2y'  =  44|  14.    4(a;  +  y)=3:ry          ) 
2  ^r'^  -  :ry  +  y'  =3  1 6  j  a:  +  y  +  a;*^  +  y'  =  26  } 

7.  x^-^xy-\^  =  ^^  15.    4:^'  +  :ry  +  4y'-=58| 
:ry-y2-2  =  0    J  5:i;'  +  5y'^  =  65           J 

8.  x'^ 
Zx 


-xy-\-y'=l  jl6.    :ry(^  +  y)=30) 

'+13a;y  +  8y'-162j  a;'  +  y''  =  35       J 


Exercise  103. 

Solve : 

1.  x  —  y  =  l  I            5.    2a;-5y  =  9        | 

2.  a;'  +  a:y  =  35j  6.    2;-y  =  9    | 
:ry-y'=6    j  ^y  +  g  =  0  j 

3.  a;y-12  =  0|  7.    5a;-7y  =  0 
^-2y  =  5   /  5:^-l^  =  4-7y' 

4.  a:y-7  =  0    j  ^ 


SIMULTANEOUS    QUADRATIC    EQUATIONS. 


267 


==1  I 


8.  x  —  y 

9.  a;'  +  4a;y  =  3    | 
4a;y  +  y^  =  2iJ 

10.  a;'  — a;y  +  3/'  =  48| 
a;  — y-8  =  0        3 

11.  a;'  +  3a;2/  +  2/2^1       ) 

12.  x'-2a;y  +  3y'  =  l-|| 


13.  x-\-y  =  a  ) 
4a;y  —  a'  =  — 46^  3 

14.  x  —  y  =  \ 
--{-^=24. 

y    X 

15.  a;'  +  9a;y  =  340) 
7a;y-y2=-17l3 

16.  a;  +  y  =  6      ) 

a;'  +  y'=723 

17.  3^:2/ +  2a; +  2 
3a;-2y  =  0 


18. 


19.    a;' +  2/' =2728 


28         ) 
-=124  3 


20. 


21. 


x''  —  xy-\'  y" 
x-^y  =  a    I 


3a;'- 4a;?/ +  5^=9 


22. 


23. 


24. 


a:4-y  r^  — .v_lQ 

a;-y      .r  +  y       3 

a;2  +  y2^45 


a;     y 
1       .       1 


17 

12 


a;+l     2/  +  1 

+  y^  =  7      ) 

'  +  /-1333 


a?  —  xy 

x^ + ^y 


25.    a; +  2/ =  4      ) 
a;*  +  /  =  823 


26.    a;' 

X- 


a     3 


27.    a;' —  a;y  =  a' 4- 6 
■2/'  =  2a6 

y'  =  4  a6  ) 
-a^-h''    3 

! 


28.    a;' 
xy 


29.    a;y  =  0 
485|  ^  +  y'=-16 

I     30.    a;'  +  a;y  +  y'^  =  37 

a;*  +  .T'y  +  /  =  481 


n.    0^  =  ax-\-hy\ 
._'    3/^  =  ay  +  5a;  3 

a;_y_2=0 
15(a;'-y')  =  16a:y 


32.    a;-y  — 2=0  ) 


33.  1±JLj^^.^zJLJ^ 
x  —  y     x-\-y     40 

6a;  =  20y  +  9 


/ 


/ 


268 


ALGEBRA. 


34. 


5+1=11 

a      b 

35.  X'  -j-y  =  ( -t  xy 

L       x'-\-y'  =  Qxy- 

2+*  =  4 

X     y         J 

36.    a;^-y^  =  3093" 

37.    f 

{X- 

-1)- 

-K^  +  i)(2/-i)  =  -iil 

i(y  +  2)  =  i(r.+  2) 

38.  10x''+15a;y  =  3a5-2aM 
10y^  +  15.ry==3a5-2^M 

39.  (2:i;  +  3y)'-2(2a;  +  33/)--8 
x'-y''  =  2l 


1  40* 


^  +  y+V^  +  ?/  =  a)     42.    a;^  +  y^  +  a;  +  y  ^  48 
--'^    :u  —  y  +  Vo;  —  y  =  b)  xy  =12 

41.    a;*-a;y4-y*=13")         43.    x' -^ xyj{- y' =  a^ 


a;'  -  a;y  +  y'  =  3 


a;  +  V:ry4-y 


44.  (a;-y)'-3(^-y)  =  10| 
a;y— 3a7y  =  54  j 

45.  -.  Va;  —  Vy  --=  ^^  (a;^  +  y^)  ) 
VJ(:r  +  y)^  =  2(a;-y)^  J 

(^  +  y) 


a;y 


54 


47.    a;  +  y+ Va;y  =  28') 
^H  y'  +  ^y  =  336  J 


Exercise  104. 


If  the  length  and  breadth  of  a  rectangle  were  each 
increased  by  1,  the  area  would  be  48 ;  if  they  were 
each  diminished  by  1,  the  area  would  be  24,  Find 
the  length  and  breadth. 


SIMULTANEOUS    QUADRATIC   EQUATIONS.  269 

2.  The  sum  of  the  squares  of  the  two  digits  of  a  number 

is  25,  and  the  product  of  the  digits  is  12.  Find  the 
number. 

3.  The  sum,  the  product,  and  the  difference  of  the  squares 

of  two  numbers  are  all  equal.     Find  the  numbers. 
Note.     Represent  the  numbers  hj  x  +  y  and  x  —  y,  respectively. 

4.  The  difference  of  two  numbers  is  f  of  the  greater,  and 

the  sum  of  their  squares  is  356.  What  are  the 
numbers? 

5.  The  numerator  and  denominator  of  one  fraction  are 

each  greater  by  1  than  those  of  another,  and  the 
sum  of  the  two  fractions  is  1^ ;  if  the  "numerators 
were  interchanged  the  sum  of  the  fractions  would 
be  1^.     Find  the  fractions. 

6.  A  man  starts  from  the  foot  of  a  mountain  to  walk  to 

its  summit.  His  rate  of  walking  during  the  second 
half  of  the  distance  is  ^  mile  per  hour  less  than  his 
rate  during  the  first  half,  and  he  reaches  the  summit 
in  5^  hours.  He  descends  in  3j  hours,  by  walking 
1  mile  more  per  hour  than  during  the  first  half  of 
the  ascent.  Find  the  distance  to  the  top  and  the 
rates  of  walking. 
Note.  Let  2x=  the  distance,  and  y  miles  per  hour  ==  the  rate  at  first. 
Then    2  +  -^  =  5i  hours,  and  -^  =  3|  hours. 

y    y-h  2/  +  1 

7.  The  sum  of  two  numbers  which  are  formed  by  the 

same  two  digits  in  reverse  order  is  -ff  of  their  dif- 
ference ;  and  the  difference  of  the  squares  of  the 
numbers  is  3960.     Determine  the  numbers. 

8.  The  hypotenuse  of  a  right  triangle  is  20,  and  the  area 

of  the  triangle  is  96.     Determine  the  sides. 
Note.     The  square  on  the  hypotenuse  =  sum  of  the  squares  on  the 
legs  ;  and  the  area  of  a  right  triangle  =  |  product  of  legs. 


270  ALGEBRA. 

9.  Two  boys  run  in  opposite  directions  round  a  rectan- 
gular field,  the  area  of  which  is  an  acre ;  they  start 
from  one  corner  and  meet  13  yards  from  the  oppo- 
site corner ;  and  the  rate  of  one  is  f  of  the  rate  of 
the  other.     Determine  the  dimensions  of  the  field. 

10.  A,  in  running  a  race  with  B,  to  a  post  and  back,  met 

him  10  yards  from  the  post.  To  make  it  a  dead 
heat,  B  must  have  increased  his  rate  from  this  point 
41|-  yards  per  minute ;  and  if,  without  changing  his 
pace,  he  had  turned  back  on  meeting  A,  he  would 
have  come  4  seconds  after  him.  How  far  was  it  to 
the  post  ? 

Note.    If  2  a;  =  the  number  of  yards  to  the  post  and  back,  and 

a;  — 10 
V  the  number   of  yards  A  runs   a  minute,   then of  y,  or 

^ — i^  =  the  number  of  yards  B  runs  a  minute, 
x  +  lO  ^ 

11.  The  fore  wheel  of  a  carriage  turns  in  a  mile  132  times 

more  than  the  hind  wheel ;  but  if  the  circumferences 
were  each  increased  by  2  feet,  it  would  turn  only 
88  times  more.     Find  the  circumference  of  each. 

12.  A  person  has  $6500,  which  he  divides  into  two  parts 

and  loans  at  different  rates  of  interest,  so  that  the  two 
parts  produce  equal  returns.  If  the  first  part  had 
been  loaned  at  the  second  rate  of  interest,  it  would 
have  produced  $180;  and  if  the  second  part  had 
been  loaned  at  the  first  rate  of  interest,  it  would 
have  produced  $245.     Find  the  rates  of  interest. 


CHAPTER  XXI. 

PROPERTIES  OF  QUADRATICS. 

263.   Every  affected  quadratic  can  be  reduced  to  the  form 
acc^  -}-  bx  -\-  c  ==  0,  the  solution  of  which  gives  the  two  roots 


L  j^ and  —'^ 

2a  2a  2a  2a 


Character  of  the  Roots. 

264.  As  regards  the  character  of  the  two  roots,  there  are 
three  cases  to  be  distinguished. 

I.  If  b^  — 4ao  is  Positive  and  not  Zero.  In  this  case  the 
roots  are  real  and  unequal.  The  roots  are  real,  since  the 
square  root  of  a  positive  number  can  be  found  exactly  or 
approximately.  If  b"^  —  ^ac  is  a  perfect  square,  the  roots 
are  rational ;  if  5^^  —  4a<?  is  not  a  perfect  square,  the  roots 
are  surds. 


The  roots  are  unequal,  since  ^ b^  —  4:ac  is  not  zero. 

II.   If  V  — 4aG  is  Zero.     In  this  case 
-eal  and  equal,  since  they  both  become  — 


II.   If  V  — 4aG  is  Zero.     In  this  case  the  two  roots  are 

b_^ 
2a 


III.  If  V-  4ac  is  Negative.  In  this  case  the  roots  are 
imaginary,  since  they  both  involve  the  square  root  of  a 
negative  number. 

The  two  imaginary  roots  of  a  quadratic  cannot  be  equal, 
since  V^  —  ^ac  is  not  zero.     They  have,  however,  the  same 


272  ALGEBRA. 

real  part,  —  — ,  and  the  same  imaginary  parts,  but  with 

opposite  signs ;  such  expressions  are  called  conjugate  imagi- 
naries.  The  expression  Z>^  —  4  ac  is  called  the  discriminant 
of  the  expression  ax^ -^hx-\-c. 

265.  The  above  cases  may  also  be  distinguished  as  follows : 

Case     I.   }?  —  ^ac> 0,  roots  real  and  unequal. 
Case    II.   5^  —  4a<?  =  0,  roots  real  and  equal. 
Case  III.   6^  —  4  ac  <  0,  roots  imaginary. 

266.  By  calculating  the  value  of  5^  —  4  ac,  we  can  deter- 
mine the  character  of  the  roots  of  a  given  equation  without 
solving  the  equation. 

(1)  a;''-5a;  +  6=-0. 

Here  a  =  ],&  =  _5^c  =  6. 

62-4ac  =  25-24  =  l. 

The  roots  are  real  and  unequal,  and  rational. 

(2)  3a:^-f  7rr-l  =  0. 

Here  a  =  3,  i  =•  7,  c  =  —  1. 

&2-4ac  =  49  + 12  =  61. 

The  roots  are  real  and  unequal,  and  are  both  surds. 

(3)  4a;2- 12:^  +  9  =  0. 

Here  a  =  4,  6  =  -  12,  c  =  9. 

62 -4ac  =  144-144  =  0. 

The  roots  are  real  and  equal. 

(4)  207^-3:^  +  4  =  0. 

Here  «  =  2,  6  =  -  3,  c  =  4. 

62-4ac  =  9-32  =  -23. 

The  roots  are  both  imaginary. 


PROPERTIES    OF    QUADRATICS.  273 

(5)    Find  the  values  of  m  for  which  the  following  equa- 
tion has  its  two  roots  equal : 

2mx''  +  (5m  +  2)^  +  (4m  +  1)  =  0. 

Here  a  =  2m,  6  =  5m  +  2,  c  =  4m  +  1. 

If  the  roots  are  to  be  equal,  we  must  have 

b^-4:ac  =  0,  or  (5m  +  2)"^  -  8m(4m  +  1)  =  0. 

2 
This  gives  m  =  2,  ©r  —  -• 

For  these  values  of  m  the  equation  becomes 

4a;2  +  12a;  +  9  =  0,  and  4^2  _  4a;  +  1  =  0, 
each  of  which  has  its  roots  equal. 

Exercise   105. 

Determine  without  solving  the  character  of  the  roots  of 
each  of  the  following  equations  : 

1.  rr' -7:r  +  12  =  0.  6.  x' +  4:X -[-1  =^0. 

2.  a:^- 7a: -30  =  0.  7.  a;^  -  2a:  + 9  -  0. 

3.  a^''^- 4a;- 5  =  0.  8.  3a;' —  4a;  —  4  =  0. 

4.  5a:'  +  8  =  0.  9.  a;' +  4a;-h4  =  0. 

5.  7a;"^- 3a; -22  =  0.  10.  7a;- 3a;^  -  2  =  0. 

Determine  the  values  of  m  for  which  the  two  roots  of 
each  of  the  following  equations  are  equal : 

^'    11.  (m  +  l)a;'  +  (m-l)a;  +  w  +  l  =  0. 

12.  (3m+l)a;^+(2m  +  2)a;  +  w  =  0. 

13.  (m-2)a;'  +  (m-5)a;  +  2m-5  =  0. 

14.  2ma;'  +  a;'  — 6ma;  — 6a;4-6m  +  l  =  0. 

15.  ma;'  +  2a;''  +  2m  =  3ma;-9a;+10. 


274  ALGEBRA. 

Relations  of  Roots  and  Coefficients. 

267.  Consider  the  equation  a;^  —  10  a;  -}-  24  =  0.  Resolve 
into  fack)rs,  {x  —  ^)(x  —  4)  =  0.  The  two  values  of  x  are 
6  and  4 ;  their  sum  is  10,  the  coefficient  of  x  with  its  sign 
changed  ;  their  product  is  24,  the  third  term. 

268.  In  general,  representing  the  roots  of  the  quadratic 
equation  aoc^  -\-hx-\-c=^0  hy  ri  and  r^,  we  have  (§  263), 


2a  2a 


b 

■Vb'-Aac 

"'-     2a~ 

2a 

Adding, 

n  +  r,  =  --; 

multiplying. 

c 
^  ^      a 

If  we  divide  the  equation  ax"^  -\-bx-\-c  =  0  through  by 

a,  we  have  the  equation  x"^  -}--x-\--  =  0]    this  may  be 

a        a 

written  x'^  -{- px  -{-  q  =  0  where  p=-,  9  =  -- 

a  a 

It  appears,  then,  that  if  any  quadratic  equation  is  made 

to  assume  the  form  x^ -{-px -}-  q  =  0,  the  following  relations 

hold  between  the  coefficients  and  roots  of  the  equation  : 

(1)  The  sum  of  the  two  roots  is  equal  to  the  coefficient 
of  X  with  its  sign  changed. 

(2)  The  product  of  the  two  roots  is  equal  to  the  constant 
term. 

Thus,  the  sum  of  the  two  roots  of  the  equation 


is  7,  and  the  product  of  the  roots  8. 


PROPERTIES   OF   QUADRATICS.  275 

269.  Eesolution  into  Pactors.     By  §  268,  if  Ti  and  rj  are 

the   roots  of  the  equation  cc^  -\~px  -\-  q—0,  the   equation 
may  be  written 

x^  —  (ri  -\-r^x-\-  r^r^  =  0. 

The  left  member  is  the  product  of  a;  —  ri,  and  x  —  rj,  so 
that  the  equation  may  be  also  written 

(x  —  r^{x~r^  =  ^. 

It  appears,  then,  that  the  factors  of  the  quadratic  expres- 
sion x^  -\-px  +  q  are  x  —  r^^  and  x  —  r^,  where  r^  and  r^  are 
the  roots  of  the  quadratic  equation  x^  -\-  px  +  g'  =  0. 

The  factors  are  real  and  different,  real  and  alike,  or 
imaginary,  according  as  r^  and  rj  are  real  and  unequal, 
real  and  equal,  or  imaginary. 

If  r^  =  Ti,  the  equation  becomes  (x  —  ri)(x  —  rj  =  0,  or 
(x  —  riy  =  0;  if,  then,  the  two  roots  of  a  quadratic  equa- 
tion are  'equal,  the  left  member,  when  all  the  terms  are 
transposed  to  that  member,  will  be  a  perfect  square  as 
regards  x. 

270.  If  the  equation  is  in  the  form  ax^  -{-bx-\-  c  =  0,  the 
left  member  may  be  written 


alx'^-\--x-i--\ 
\        a        aj 

a{x~  ri){x  —  r^. 


271.    If  the  roots  of  a  quadratic  equation  are  given,  we 
can  readily  form  the  equation. 

5 

Form  the  equation  of  which  the  roots  are  3  and  —  -  • 


The  equation  is      (a;  —  3)[  x  +  -  j=  0, 

or  (x-3)(2a;  +  5)  =  0, 

or  2x2 -a -15  =  0. 


276  ALGEBRA. 

272.  Any  quadratic  expression  may  be  resolved  into 
factors  by  putting  the  expression  equal  to  zero,  and  solving 
the  equation  thus  formed. 

(1)  Resolve  into  two  factors  x^  —  bx-\-S. 
Write  the  equation 

Solve  this  equation,  and  the  roots  are  found  to  be 

5+  VT3  „„,  5-V13 

~^—  ^^^  -T— 
Therefore,  the  factors  of  x'^  —  5  a?  +  3  are 

^     5+  Vl3  ^.^     5  -  Vl3 

X and  X 


(2)  Resolve  into  factors  Sx^ —  4:X-\~5. 
Write  the  equation 

Solve  this  equation,  and  the  roots  are  found  to  be 

3  3 

Therefore,  the  expression  Sx"^  —  4:X  +  5  may  be  written  (§  270), 


K^-^-^^)e--t^) 


Exercise  106. 
Form  the  equations  of  which  the  roots  are 

7.  a-2b,  Sa  +  2b. 

8.  2a  ~b,  b~Sa. 

9.  a-fl,  1  —  a. 

13.  x'-Sx  +  i. 

14.  x'-^-x-i-l. 

15.  4x'i-12x-j-13. 


1. 

2,1. 

4.    I 

-i 

2. 

7, - 

3. 

5.    - 

5,  ~ 

f 

3. 

ii 

6.    - 

if 

Resolve  into  factors : 

10. 

3a:'- 

-15a:- 

-42. 

11. 

9a:'- 

-27a:- 

-70. 

12. 

49  a:' 

-f49a; 

+  6. 

CHAPTER   XXII. 
RATIO,  PROPORTION,   AND  VARIATION. 

273.  Eatio  of  Numbers.  The  relative  magnitude  of  two 
numbers  is  called  their  ratio  when  expressed  by  the  indi- 
cated quotient  of  the  first  by  the  second.     Thus,  the  ratio 

of  a  to  5  is  -,  ov  a~h,  ov  a-.h, 

0 

The  first  term  of  a  ratio  is  called  the  antecedent,  and  the 
second  term  the  consequent.  When  the  antecedent  is  equal 
to  the  consequent,  the  ratio  is  called  a  ratio  of  equality ; 
when  the  antecedent  is  greater  than  the  consequent,  the 
ratio  is  called  a  ratio  of  greater  inequality ;  when  less,  a  ratio 
of  less  inequality. 

When  the  antecedent  and  consequent  are  interchanged, 
the  resulting  ratio  is  called  the  inverse  of  the  given  ratio. 
Thus,  the  ratio  3  :  6  is  the  inveise  of  the  ratio  6  :  3. 

274.  A  ratio  will  not  be  altered  if  both  its  terms  are 
multiplied   by  the  same  number.      For  the  ratio  a-:h  is 

represented  by  -,  the  ratio  ma  :  mb  is  represented  by  ^^ ; 
o  mb 

and  since  — -  =  -,  we  have  ma  :  rtih  =  a:h. 
mb      b 

275.  A  ratio  will  be  altered  if  difierent  multipliers  of  its 
terms  are  taken ;  and  will  be  increased  or  diminished  ac- 
cording as  the  multiplier  of  the  antecedent  is  greater  than 
or  less  than  that  of  the  consequent.     Thus, 


278 

ALGEBRA. 

If 

W>  W, 

If 

m<n, 

then 

ma  >  na, 

then 

ma  <  na , 

and 

ma     na , 
nb      nb  ' 

and 

7na  ^  na , 
nb       nb  ' 

but 

na_a 
nb~b 

but 

na     a 
nb~b 

ma      a 
"  nb       b' 

.    TYia      a 
'  '   nb       b 

or 

ma  :  nb  ^  a  :  b. 

or 

ma  -.nbKa-.b 

276.  Ratios  are  compounded  by  taking  the  product  of  the 
fractions  that  represent  them.  Thus,  the  ratio  compounded 
oi  a:b  and  c:di^  ac\  bd. 

The  ratio  compounded  of  a  :  5  and  a  :  5  is  the  duplicate 
ratio  a^ :  h^ ;  the  ratio  compounded  of  a  :  5,  a  :  b,  and  a  :  b 
is  the  triplicate  ratio  a^  -.b^;  and  so  on. 

277.  Ratios  are  compared  by  comparing  the  fractions 
that  represent  them. 

a  :  5  >  or  <  c  :  c?, 


Thus, 
according  as 

as 

as 


a  ^         .c 
ad  ^        ^bc 
ad  >  or  <  be. 


278.  Proportion  of  Numbers.  Four  numbers,  a,b,c,  d,  are 
said  to  be  in  proportion  when  the  ratio  a:b  is  equal  to  the 
ratio  c :  d. 

We  then  write  a  :  b  =  c  :  d,  and  read  this,  the  ratio  of  a 
to  b  equals  the  ratio  of  c  to  d,  or  a  is  to  5  as  c  is  to  d. 

A  proportion  is  also  written  a  :  b  :  :  c  :  d. 

The  four  numbers,  a,  b,  c,  d,  are  called  proportionals ;  a 
and  d  are  called  the  extremes,  b  and  c  the  means. 


RATIO   AND    PROPORTION.  279 

279.    When  four  numbers  are  in  proportion,  the  product 
of  the  extremes  is  equal  to  product  of  the  means. 

For,  if  a:h  =  c  :  d, 

then  -  =  -' 

b      d 

Multiplying  by  hd,       ad  =  be. 

The  equation  ad  =  be  gives  a  =  —,  b=  —  ;  so  that  an 

d  G 

extreme  may  be  found  by  dividing  the  product  of  the 
means  by  the  other  extreme  ;  and  a  mean  may  be  found  by 
dividing  the  product  of  the  extremes  by  the  other  mean. 
If  three  terms  of  a  proportion  are  given,  it  appears  from 
the  above  that  the  fourth  term  can  have  one,  and  but  one, 
value. 


280.  If  the  product  of  two  numbers  is  equal  to  the 
product  of  two  others,  either  two  may  be  made  the  ex- 
tremes of  a  proportion  and  the  other  two  the  means. 

For,  if  ad  =  be, 

then,  dividing  by  5c?,        ad _  be 

bd~bd 


or 


a__£ 
b~d 

a  :  b  =  c  :  d. 


281.   Transformations  of  a  Proportion.     If  four  numbers,  a, 
b,  c,  d,  are  in  proportion,  they  will  be  in  proportion  by : 

I.   Inversion ;  that  is,  b  will  be  to  a  as  c?  is  to  e. 
F«r,  if  a:b  =  c:d, 

then  ?  =  4. 

b     a 


280  ALGEBRA. 


1  -,      a      -,       c 

and  1  -^-  -  =  i  -f-  -. 


b  d 

h      d 


or 

a     c 


.'.  h  :a  =  d:c. 

II,   Composition ;  that  is,  a  +  5  will  be  to  6  as  c+ c?  is  to  d. 
For,  if  a:h  =  c  :  d. 


th 


en 


b~  d 


and  ^+1=^+1, 


c?  is  to  d. 
For,  if 

then 
and 

b      d 

:.  a—h\h  —  c  —  d\d. 

IV.   Composition  and  Division;  that  is,  a-^h  will  be  to 
a  —  6  as  c  -f  c?  is  to  c  —  c?. 

a-\-h      c-\-  d 


h 

d 

+  ^: 

:b-- 

=  c-^d: 

c^. 

is,  a 

—  b  will  be 

to 

a\ 

:i: 

=  c:d, 

a 

c 

b 

d 

a 

•1    : 

=  --1, 

b 

d       ' 

• 

a- 

1^- 

_c-(^ 

For,  from  II., 
and  from  III., 
Dividing, 


b  d 

a  —  b      c  —  d 


b  d 

a-\-b c  -\-  d 

a  —  b      c  —  d 
.  a-\-b'.a  —  b  =  c-{-d:c  —  d. 


RATIO   AND   PROPORTION.  281 

V.   Alternation ;  that  is,  a  will  be  to  c  as  6  is  to  d. 


For,  if 

a  \b  =  c  \  d, 

then 

a      c 
b~  d 

Multiplying  by  -» 
c 

ah  __bc 
be      cd 

or 

a      b 
c      d 

\  a  \  c  =  h  :  d. 

282.  In  a  Series  of  Equal  Eatios,  the  sum  of  the  antecedents 
is  to  the  sum  of  the  consequents  as  any  antecedent  is  to  its 
consequent. 

-ci       -e         OL      c      e      q 
T  may  be  put  for  each  of  these  ratios. 

rty\  OL  c  e  q 

Then  -  =  r,  -  =  r,  -  =  r,  --f-r. 

h  d         J  h 

:.  a^=br,  c  —  dr,  e=fr,  g=^hr. 
.-.  a  +  c  +  e-\-g^{b-\-d-\-f-^h)r. 
•    (^  +  c-\-e-\-g  _^_a 
"  b  +  d-]-f+h  b 

.'.  a  +  ci-e  +  gib  +  d  +/+  h  =  a:b. 
In  like  manner  it  may  be  shown  that 

ma-{-nc-\-pe-\-qg  :  mb-\-nd-^pf-{-qh  =  a:b. 

283.  Continued  Proportion.  Numbers  are  said  to  be  in 
continued  proportion  when  the  first  is  to  the  second  as  the 
second  is  to  the  third,  and  so  on.     Thus,  a,  6,  c,  d,  are  in 

continued  proportion  when  t  —  ~  ~  ~i' 


282  ALGEBRA. 

284.  If  a,h,c  are  proportionals,  so  that  a:h  =  h  :  c,  then 
h  is  called  a  mean  proportional  between  a  and  c,  and  c  is 
called  a  third  proportional  to  a  and  h. 

If  a  :  b  =  b  :  c,  then  6  =  Vac 

For,  if  a  :  b  =  b  :  c, 

then  T  — "' 

and  b^  —  ac. 

.'.  ^  =^ac. 

285.  The  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 

For,  if  a  :  b  —  c    :  d, 

e:f=g   :  A, 

k  :  I  ^=  7)1 :  n, 

,1  a      c     e      q     h      m 

then  T^~7'    -r^i'    7^  — 

b      a    J      h     L      n 

.  Taking  the  product  of  the  left  members,  and  also  of  the 
right  members  of  these  equations, 

aek  _  cgm 

bfl      dhn 

.'.  aek  :  bfl  =  cgm  :  dhn. 

286.  Like  powers,  or  like  roots,  of  the  terms  of  a  propor- 
tion are  in  proportion. 

For,  if  a  .  b  =  c  \  d, 

then  T  —  --,- 

b      d 

Kaising  both  sides  to  the  nth  power, 

b""      d""' 
.-.  a** :  5"  =  c** :  c^^ 


RATIO   AND   PROPORTION.  263 


Extracting 

the  nth  root, 

1 

a" 

1  ' 

1 

r 
1 

I 

.'.  a" 

:5«^ 

=  c"; 

:c?" 

287.  The  laws  that  have  been  established  for  ratios 
should  be  remembered  when  ratios  are  expressed  in  frac- 
tional form. 

(1)   Solve  :  ^l±^±l ^ ^^-^  +  2 

^  ^  x^-x-l      x^-\-x-2 

By  composition  and  division, 

2x^  2x^ 


2(aj  +  l)     -2{x-2) 

This  equation  is  satisfied  when  x  =  0.     For  any  other  value  of  x, 
we  may  divide  by  x^. 

We  then  have 


x  +  1      2  —  x 
and  therefore  x  =  ^. 

(2)   li  a  :  h  =  G  :  d,  show  that 

o? -\- ah  :  h^  —  ah  =  c^  -{-cd-.d^  —  cd. 

If 

then 
and 

that  is, 


a      c 
h~  d 

a  +  b      c  +  d 
a—b      c  —  d 

a         c 
-h      -d 

a 

a  +  6        c    v^  +  ^. 
a  —  6      —  d     c  —  d 

a?  +  a&      c2  4-  erf 

&2  _  a&      rf2  _  c^; 

a?  +  ah 

:  62  ^  a6  =  c2  +  erf :  rf2  _  erf. 

284  ALGEBRA. 

(3)    U  a  :  b  =  c  :  d,  and  a  is  the  greatest  term,  show  that 
a-}-  d  is  greater  than  b  -\-  c. 


Since 

a      c        t     ^ 
-  =  -,  and  a>c, 
b      d 

(1) 

the  denominator 

h>d. 

From  (1),  by  division, 

a—b      c - d 
b            d 

(2) 

Since 

b>d, 

from  (2), 

a  —  b'>c—d. 

Now, 

b+d=b  +  d. 

Adding, 

a  +  d>b  +  c. 

288.  Ratio  of  Quantities.  To  measure  a  quantity  of  any- 
kind  is  to  find  out  how  many  times  it  contains  another 
known  quantity  of  the  same  kind,  called  the  unit  of  measure. 
Thus,  if  a  line  contains  5  times  the  linear  unit  of  measure, 
one  yard,  the  length  of  the  line  is  5  yards. 

289.  Commensurable  Quantities.  If  two  quantities  of  the 
same  kind  are  so  related  that  a  unit  of  measure  can  be 
found  which  is  contained  in  each  of  the  quantities  an  in- 
tegral number  of  times,  this  unit  of  measure  is  a  comm-on 
7neasure  of  the  two  quantities,  and  the  two  quantities  are 
said  to  be  commensurable. 

If  two  commensurable  quantities  are  measured  by  the 
same  unit,  their  ratio  is  simply  the  ratio  of  the  two  numbers 
by  which  the  quantities  are  expressed.  Thus,  -J-  of  a  foot  is 
a  common  measure  of  2|-  feet  and  3f  feet,  being  contained 
in  the  first  15  times  and  in  the  second  22  times. 

The  ratio  of  2|-  feet  to  3-|  feet  is  therefore  the  ratio  of 
15  :  22. 

Evidently  two  quantities  different  in  kind  can  have  no 
ratio. 


RATIO   AND   PROPORTION.  285 

290.  Incommensurable  Quantities.  The  ratio  of  two  quan- 
tities of  the  same  kind  cannot  always  be  expressed  by  the 
ratio  of  two  whole  numbers.  Thus,  the  side  and  diagonal 
of  a  square  have  no  common  measure ;  for  if  the  side  is  a 
inches  long,  the  diagonal  will  be  a V2  inches  long,  and  no 
measure  can  be  found  which  will  be  contained  in  each  an 
integral  number  of  times. 

Again,  the  diameter  and  circumference  of  a  circle 
have  no  common  measure,  and  are  therefore  incommen- 
surable. 

In  this  case,  as  there  is  no  common  measure  of  the  two 
quantities,  we  cannot  find  their  ratio  by  the  method  of 
§  289.     We  therefore  proceed  as  follows  : 

Suppose  a  and  5  to  be  two  incommensurable  quantities 
of  the  same  hind.  Divide  h  into  any  integral  number  (n) 
of  equal  parts,  and  suppose  one  of  these  parts  is  contained 
in  a  more  than  m  times  and  less  than  m  +  I  times.     Then 

the  ratio  7>— ,  but  <^ "^     ;  that  is,  the  value  of  7  lies 
0      n  n  0 

between  ™  and  ''!^^+^- 
n  n 

The  error,  therefore,  in  taking  either  of  these  values  for 

-  is  less  than  —     But  by  increasing  n  indefinitely,  _  can 

be  made  to  decrease  indefinitely,  and  to  become  less  than 
any  assigned  value,  however  small,  though  it  cannot  be 
made  absolutely  equal  to  zero. 

Hence,  the  ratio  of  two  incommensurable  quantities  can- 
not be  expressed  exactly/  in  figures,  but  it  may  be  expressed 
approximately  to  any  desired  degree  of  accuracy. 

Thus,  if  b  represents  the  side  of  a  square,  and  a  the 
diagonal, 

7  =  V2. 


286  ALGEBRA. 

Now  V2  =  1.41421356 ,  a  value  greater  than  1.414213, 

but  less  than  1.414214. 

If  then,  a  millionth  part  of  b  is  taken  as  the  unit,  the 

1        ^  ^x.        r    «  r      k  4-  1414213        .   1414214 

value  of  the  ratio  -  lies  between  j^^^^  and  ^^^^. 

and  therefore  differs  from  either  of  these  fractions  by  less 

than  • 

1000000 

By  carrying  the  decimal  farther,  a  fraction  may  be  found 
that  will  differ  from  the  true  value  of  the  ratio  by  less  than 
a  billionth,  a  trillionth,  or  any  other  assigned  value  what- 
ever. 

291.  The  ratio  of  two  incommensurable  quantities  is  an 
incommensurable  ratio,  and  is  a  fixed  value  toward  which  its 
successive  approximate  values  constantly  tend  as  the  error 
is  made  less  and  less. 

292.  Proportion  of  Quantities.  In  order  for  four  quanti- 
ties, A,  B,  C,  D,  to  be  in  proportion,  A  and  B  must  be  of 
the  same  hind,  and  C  and  D  of  the  same  kind  (but  C  and 
D  need  not  necessarily  be  of  the  same  kind  as  A  and  ^), 
and  in  addition  the  ratio  of  ^  to  .5  must  be  equal  to  the 
ratio  of  G  to  D, 

If  this  be  true,  we  have  the  proportion 

A.B=^C\D. 

When  four  quantities  are  in  proportion,  the  numbers  by 
which  they  are  expressed  are  four  abstract  numbers  in 
proportion. 

293.  The  laws  of  §  281,  which  apply  to  proportion  of 
numbers,  apply  also  to  proportion  of  quantities,  except  that 
alternation  will  apply  only  when  the  four  quantities  in 
proportion  are  all  of  the  same  kind. 


RATIO   AND   PROPORTION.  287 

Exercise  107. 
If  a  :  b  :  :  c  :  d,  prove  that : 

1.  7na  :nb  ::  mc  :  nd.  4.    a^ :  b^ :  :  c^ :  d^. 

2.  Sa-\-b:b  ::Sc-{-  d:d,      5.    a:  a  + b  :  :  c  :  c-\- d. 

3.  a-{-2b  :b  :  :  c-{-2d:  d.      6.    a  :  a~b  :  :  c  :  c  —  d. 

7.  ma  +  nb  :  ma  —  nb:\  mc  +  wc? :  9?i<?  —  nd. 

8.  2a  +  3^»:3a-45::2c  +  3c?:3c-4£Z. 

9.  ma^  +  nc^ :  mJ^  +  w^^  ::«'*:  51 

10.    wa'^  +  nab  -{-pb"^ :  7?2c'^  +  ncd  -\-pd^ :  :  b'^ :  d^. 

If  a  :  b  :  :  b  :  c,  prove  that : 

11.    ai-b:b-{-c::a:b.  12.    a""  +  ab  :  b"" -{-be  :  :  a:  c. 

13.    a  :  c  :  :  (a  +  bf  :  (5  +  c)l 

14.  When  a,  b,  and  c  are  proportionals,  and  a  the  greatest, 

show  that  a-\-c>2b. 

15.  If  — -Ji-  =  il—.= ,  and  x,  y,  z  are  unequal,  then 

I  'in  n 

l-{-m-\-n  —  0. 

16.  Find  a;  when  a;  + 5  :  2a:  -  3  :  :  5.^  +  1  :  3a;  — 8. 

17.  Find  x  when  x  -}-  a  :  2x  —  b  :  :  3 x  -\-  b  :  4iX  —  a. 

18.  Find  X  when  Vrr  +  V6  :  ^/x  —  Vb  :  :  a:b. 

19.  Find  X  and  y  when  a; :  27  :  :  y  :  9,  and  a; :  27 : :  2 :  a:— y. 

20.  Find  X  and  3/  when  x -{-y -\-l  :  x-{-y -\-2:  :  6  :7,  and 

when  2/4-2a;:2/  — 2rr:  :  12a:+6y  — 3  :  63/  — 12ar— 1. 

21.  Find  x  when  a;''-4a:-K2  :  a;^-2x-l  : :  a:'-4a: :  a;'-2a;-2. 


288  ALGEBRA.     " 

22.  A  railway  passenger  observes  that  a  train  passes  him, 

moving  in  the  opposite  direction,  in  2  seconds ;  but 
moving  in  the  same  direction  with  him,  it  passes  him 
in  30  seconds.    Compare  the  rates  of  the  two  trains. 

23.  A  and  B  trade  with  different  sums.    A  gains  1 200  and 

B  loses  $50,  and  now  A's  stock  :  B's  :  :  2  :  -J.  But,  if 
A  had  gained  f  100  and  B  lost  $85,  their  stocks 
would  have  been  as  15  :  Sh  Find  the  original  stock 
of  each. 

24.  A  quantity  of  milk  is  increased  by  watering  in  the 

ratio  4 : 5,  and  then  3  gallons  are  sold  ;  the  remainder 
is  mixed  with  3  quarts  of  water,  and  is  increased  in 
the  ratio  6 :  7.  How  many  gallons  of  milk  were 
there  at  first  ? 

25.  In  a  mile  race  between  a  bicycle  and  a  tricycle  their 

rates  were  as  5  :  4.  The  tricycle  had  half  a  minute 
start,  but  was  beaten  by  176  yards.  Find  the  rates 
of  each. 

26.  The  time  which  an  express- train  takes  to  travel  180 

miles  is  to  that  taken  by  an  ordinary  train  as  9  :  14. 
The  ordinary  train  loses  as  much  time  from  stopping 
as  it  would  take  to  travel  30  miles ;  the  express- 
train  loses  only  half  as  much  time  as  the  other  by 
stopping,  and  travels  15  miles  an  hour  faster.  What 
are  their  respective  rates  ? 

27.  A  line  is  divided  into  two  parts  in  the  ratio  2 :  3,  and 

into  two  parts  in  the  ratio  3:4;  the  distance  be- 
tween the  points  of  section  is  2.  Find  the  length 
of  the  line. 

28.  When  a,  b,  c,  d,  are  proportional  and  unequal,  show 

that  no  number  x  can  be  found  such  that  a-{-  x, 
h  -^  X,  c-{-x,  d-\-x,  shall  be  proportionals. 


ratio  and  proportion.  289 

Variation. 

294.  A  quantity  which  in  any  particular  problem  has  a 
fixed  value  is  called  a  constant  quantity,  or  simply  a  constant ; 
a  quantity  which  may  change  its  value  is  called  a  variable 
quantity,  or  simply  a  variable. 

Variable  numbers,  like  unknown  numbers,  are  generally 
represented  by  x,  y,  z,  etc. ;  constant  numbers,  like  known 
numbers,  by  a,  &,  c,  etc. 

295.  Two  variables  may  be  so  related  that  when  a  value 
of  one  is  given,  the  corresponding  value  of  the  other  can  be 
found.  In  this  case  one  variable  is  said  to  be  a  function 
of  the  other ;  that  is,  one  variable  depends  upon  the  other 
for  its  value.  Thus,  if  the  rate  at  which  a  man  walks  is 
known,  the  distance  he  walks  can  be  found  when  the  time 
is  given  ;  the  distance  is  in  this  case  ^function  of  the  time. 

296.  There  is  an  unlimited  number  of  ways  in  which 
two  variables  may  be  related.  We  shall  consider  in  this 
chapter  only  a  few  of  these  ways. 

297.  When  x  and  y  are  so  related  that  their  ratio  is 
constant,  y  is  said  to  vary  as  x ;  this  is  abbreviated  thus : 
2/ oca:.  The  sign  oc,  called  the  sign  of  variation,  is  read 
"varies  as."  Thus,  the  area  of  a  triangle  with  a  given 
base  varies  as  its  altitude ;  for,  if  the  altitude  is  changed 
in  any  ratio,  the  area  will  be  changed  in  the  same  ratio. 

In  this  case,  if  we  represent  the  constant  ratio  by  w, 

y 
y  '.  X  —  m,  OT  ~  =  m  ;  /.  y  =  mx. 

Again,  \iy\  x'  and  y",  x"  be  two  sets  of  corresponding 
values  of  y  and  x,  then 

y'  :x'  =  y"  :  x" ; 
by  alternation,  y' :  y"  ^  x'  :  x". 


290  ALGEBRA. 

298.    When  x  and  y  are  so  related  that  the  ratio  of  y  to  - 

is  constant,  y  is  said  to  vary  inversely  as  x  ;  this  is  written 

y  oc  -.     Thus,  the  time  required  to  do  a  certain  amount  of 

work  varies  inversely  as  the  number  of  workmen  employed  ; 
for,  if  the  number  of  workmen  be  doubled,  halved,  or 
changed  in  any  ratio,  the  time  required  will  be  halved, 
doubled,  or  changed  in  the  inverse  ratio. 

that  is, 


In  this  case,  y:--=m\  •'•  3/  "=  ^.  and  xy  = 

■■  m 

the  product  xy  is  constant. 

As  before,                   y:l  =  y':l, 

X                   X 

x'y^  =  xy\ 

or                                     y'  :  y"  =  x"  :  x'. 

§280 

299.  If  the  ratio  oi  y :  xz  is  constant,  then  y  is  said  to 
vary  jointly  as  a;  and  z. 

In  this  case,  y  =  mxz, 

and  y' :  y"  =  x^z^ :  x'^z'\ 

300.  If  the  ratio  3/ :  ^  ^^  constant,  then  y  varies  directly 
as  X  and  inversely  as  z. 


In  this  case,  y  = 


7nx 


and  y:y'  =  '?^:^?^  =  ?:.^ 


301.   Theorems. 

I.    If  y  oc  a:,  and  a;  oc  2,  then  y  oc  z. 

For  y  =  mx  and  a;  =  n2;. 

.*.  y  —  mnz. 

.*.  y  varies  as  z. 


RATIO    AND    PROPORTION.  291 

II.  li  y  OCX,  and  x  ccz,  then  (y  dr  z) oc x. 
For  y  =  7nx  and  z  =  nx. 

.'.  y  ±  z  =  (771  ±  w)a;. 
.*.  3/  ±  z  varies  as  a;. 

III.  If  y  ccx  when  2;  is  constant,  and  y  ccz  when  a;  is 
constant,  then  y  ccxz  when  a;  and  z  are  both  variable. 

Let  x',  y\  z\  and  a;",  y",  2",  be  two  sets  of  corresponding 
values  of  the  variables. 

Let  X  change  from  x^  to  a:",  z  remaining  constant,  and  let 
the  corresponding  value  of  y  be  Y. 

Then  y':    Y=x':x^\  (1) 

Now  let  z  change  from  2'  to  2",  rr  remaining  constant. 

Then  Y:y^^-=--z':z'\  (2) 

From  (1)  and  (2), 

y'Y:y''Y=xh'  -x'^z^^  §285 

or  y' :  y"     =  xh^ :  x"z", 

or  y':a:V   =  y"    :  .r V.  §  281,  V. 

..".  the  ratio  -^  is  constant,  and  y  varies  as  0:2. 

In  like  manner,  it  may  be  shown  that  if  y  varies  as  each 
of  any  number  of  quantities  x,  2,  u,  etc.,  when  the  rest  are 
unchanged,  then  when  they  all  change,  y  oc  xzu,  etc. 

Thus,  the  area  of  a  rectangle  varies  as  the  base  when  the  altitude 
is  constant,  and  as  the  altitude  when  the  base  is  constant,  but  as  the 
product  of  the  base  and  altitude  when  both  vary. 

The  volume  of  a  rectangular  solid  varies  as  the  length  when  the 
width  and  thickness  remain  constant;  as  the  width  when  the  length 
and  thickness  remain  constant ;  as  the  thickness  when  the  length  and 
width  remain  constant;  but  as  the  product  of  length,  breadth,  and 
thickness  when  all  three  vary. 


202  ALGEBRA. 

302.   Examples. 

(1)  If  y  varies  inversely  as  x,  and  when  y  =  2  the  cor- 
responding value  of  X  is  36,  find  the  corresponding  value 
of  X  when  3/  =  9. 


[lere 

y- 

X 

=  xy 

.'.  m  = 

=  2  X  36  = 

72. 

■f  9  and  72 

are 

substituted  for  3/  and 

m  respectively 

in 

y  = 

_  m 

X 

result  is 

9  = 

=  ^,  or9a; 

X 

=  72. 

(2)  The  weight  of  a  sphere  of  given  material  varies  as 
its  volume,  and  its  volume  varies  as  the  cube  of  its  diam- 
eter.    If  a  sphere  4  inches  in  diameter  weighs  20  pounds, 
find  the  weight  of  a  sphere  5  inches  in  diameter. 
Let  W  represent  the  weight, 

V  represent  the  volume, 
D  represent  the  diameter. 
Then  TToc  F  and   V  oc  D^. 

.-.   WocL^.  ^301,1. 

Put  W=ml^; 

then,  since  20  and  4  are  corresponding  values  of  W  and  D, 

20  =  ?7i  X  64. 

•   -m      20      5 
. .  m  =  —  =  — 
64     16 

.-.  when  Z>  =  5,  Pr=  t\  of  125  =  39^^ 

Exercise  108. 

1.  If  ^  oc^,  and  ^  =-4  when  ^  =  5,  find  A  when  £=  12. 

2.  If  ^  oc  ^,  and  when  B  =  ^,  A  =  i,  find  ^  when  ^-i 

3.  If  yl  vary  jointly  as  B  and  C,  and  3,  4,  5,  be  simulta- 

lieous  values  of  ^,  JB,  C,  find  A  when  £  =  C  =  10. 


VARIATION.  293 

4.  If  ^  oc  -- ,  and  when  ^  =  10,  B  -=2,  find  the  value  of 

B  when  ^  =  4. 

5.  If  ^  oc  -,  and  when  ^  =  6,  5  =  4,  and  (7=3,  find 

C 

the  value  of  A  when  B  =  b  and  C=1. 

6.  If  the  square  of  X  varies  as  the  cube  of  F,  and  Jr=  3 

when  F=4,  find  the  equation  between  Xand  Y. 

7.  If  the  square  of  Xvaries  inversely  as  the  cube  of  F,  and 

X=2  when  F=  3,  find  the  equation  between  X 
andF. 

8.  If  Ovaries  as  X  directly  and  F  inversely,  and  if  when 

Z^2,  X=  3,  and  Y=  4,  find  the  value  of  Z  when 
X=15and  F=8. 

9.  li  Ace  B  -{-c  where  c  is  constant,  and  if  ^  =  2  when  B 

F=  1,  and  if  ^  =  5  when  5  =  2,  find  A  when  5  =  3. 

10.  The  velocity  acquired  by  a  stone  falling  from  rest 

varies  as  the  time  of  falling;  and  the  distance  fallen 
varies  as  the  square  of  the  time.  If  it  be  found  that 
in  3  seconds  a  stone  has  fallen  145  feet,  and  acquired 
a  velocity  of  96f  feet  per  second,  find  the  velocity 
and  distance  at  the  end  of  5  seconds. 

11.  If  a  heavier  weight  draw  up  a  lighter  one  by  means 

of  a  string  passing  over  a  fixed  wheel,  the  space 
described  in  a  given  time  will  vary  directly  as  the 
difference  between  the  weights,  and  inversely  as 
their  sum.  If  9  ounces  draw  7  ounces  through  8 
feet  in  2  seconds,  how  high  will  12  ounces  draw  9 
ounces  in  the  same  time  ? 


294  ALGEBRA. 

12.  The  space  will  vary  also  as  the  square  of  the  time. 

Find  the  space  in  Example  11,  if  the  time  in  the 
latter  case  is  3  seconds. 

13.  Equal  volumes  of  iron  and  copper  are  found  to  weigh 

77  and  89  ounces  respectively.  Find  the  w^eight  of 
10^  feet  of  round  copper  rod  when  9  inches  of  iron 
rod  of  the  same  diameter  w^eigh  olj-^-^  ounces. 

14.  The  square  of  the  time  of  a  planet's  revolution  varies 

as  the  cube  of  its  distance  from  the  sun.  The  dis- 
tances of  the  Earth  and  Mercury  from  the  sun  being 
91  and  35  millions  of  miles,  find  in  days  the  time  of 
Mercury's  revolution. 

15.  A  spherical  iron  shell  1  foot  in  diameter  weighs  ^^g- 

of  what  it  would  weigh  if  solid.  Find  the  thick- 
ness of  the  metal,  knowing  that  the  volume  of  a 
sphere  varies  as  the  cube  of  its  diameter. 

16.  The  volume  of  a  sphere  varies  as  the  cube  of  its  diame- 

ter. Compare  the  volume  of  a  sphere  6  inches  in 
diameter  with  the  sum  of  the  volumes  of  three  spheres 
whose  diameters  are  3,  4,  5  inches  respectively. 

17.  Two  circular  gold  plates,  each  an  inch  thick,  the  diam- 

eters of  which  are  6  inches  and  8  inches  respectively, 
are  melted  and  formed  into  a  single  circular  plate 
1  inch  thick.  Find  its  diameter,  having  given  that 
the  area  of  a  circle  varies  as  the  square  of  its  diameter. 

18.  The  volume  of  a  pyramid  varies  jointly  as  the  area  of 

its  base  and  its  altitude.  A  pyramid,  the  base  of 
which  is  9  feet  square,  and  the  height  of  which  is 
10  feet,  is  found  to  contain  10  cubic  yards.  What 
must  be  the  height  of  a  pyramid  upon  a  base  3  feet 
square,  in  order  that  it  may  contain  2  cubic  yards? 


CHAPTER  XXIII. 
PROGRESSIONS. 

303.  A  succession  of  numbers  that  proceed  according  to 
some  fixed  law  is  called  a  series ;  the  successive  numbers 
are  called  the  terms  of  the  series. 

A  series  that  ends  at  some  particular  term  is  a  finite 
series;  a  series  that  continues  without  end  is  an  infinite 
series. 

304.  The  number  of  different  forms  of  series  is  unlimited  ; 
in  this  chapter  we  shall  consider  only  Arithmetical  Series, 
Geometrical  Series,  and  Harmonical  Series. 

Arithmetical  Progression. 

305.  A  series  is  called  an  arithmetical  series  or  an  arith- 
metical progression  when  each  succeeding  term  is  obtained 
by  adding  to  the  preceding  term  a  constant  difference. 

The  general  representative  of  such  a  series  will  be 

a,  a-\-  d,  a-\-2d,  a-\-^d , 

in  which  a  is  the  first  term  and  d  the  common  difierence ; 
the  series  will  be  increasing  or  decreasing  according  as  d  is 
positive  or  negative. 

306.  The  nth  Term,  Since  each  succeeding  term  of  the 
series  is  obtained  by  adding  d  to  the  preceding  term,  the 
coefficient  of  d  will  always  be  one  less  than  the  number  of 
the  term,  so  that  the  nth  term  is,  a-\-(n  —  l)d. 


296  ALGEBRA. 

If  the  nth  term  is  represented  by  I,  we  have 

l=.a-Y(n-l)d.  I. 

307.  Sum  of  the  Series.  If  I  denotes  the  nth  term,  a  the 
first  term,  n  the  number  of  terms,  d  the  common  difference, 
and  s  the  sum  of  n  terms,  it  is  evident  that 

s=     a     +(a+c/)-f(«  +  2c?)  + -\-{l-d)-\-      I,    or 

s=      I     +(^ -/)  +  (/ -2c?) -f +  (a+c?)-h     a. 

—  n{a-\- 1). 
.•.s  =  |(a  +  0.       .  11. 

308.  From  the  two  equations, 

l=:a-\-{n-l)d,  I. 

s=|(«  +  0.  n. 

any  two  of  the  five  numbers  a,  c?,  Z,  w,  s  may  be  found  when 
the  other  three  are  given. 

(1)  Find  the  sum  of  ten  terms  of  the  series,  2,  5,  8,  11, 

Here  a  =  2,  d  =  \  n  =  10. 

From  I.,  ^  =  2 +  27  =  29. 

Substituting  in  IL,  s  =  —  (2  +  29)  =  155. 

(2)  The  first  term  of  an  arithmetical  series  is  3,  the  last 
term  31,  and  the  sum  of  the  series  136.     Find  the  series. 

From  I.  and  IL,  31  =  3  +  (n  -  l)c/,  (1) 

136  =  ^(3  +  31).  (2) 

From  (2),  ?i  =  8. 

Substituting  in  (1),  c?  =  4. 

The  series  is    3,     7,      11.     15,     19,     23,     27,     31. 


ARITHMETICAL   PROGRESSION.  '297 

(3)  How  many  terms  of  the  series,  5,  9,  13,  ,  must  be 

taken'  in  order  that  their  sum  may  be  275? 

From  I.,  Z  =  5  +  (n-l)4. 

.-.  Z  =  4n  +  1.  (1) 

From  II,  275  =  ^(5  +  0.  (2) 

Substituting  in  (2)  the  vah;e  of  I  found  in  (1), 

275  =  -(4n  +  6), 

or  2n2  +  3n=275. 

We  now  have  to  solve  this  quadratic. 
Complete  the  square, 

16n2  + 0  +  9  =  2209. 
Extract  the  root,  4n  +  3  =  ±  47. 

.-.  n  =  ll,  or-  12J. 
We  use  only  the  positive  result. 

(4)  Find  n  when  d,  I,  s  are  given. 
From  I.,  a  =  l  —  (n  —  l)  d. 

-p         TT  2s  —  In 

From  II.,  a  = 

Therefore,  l—{n  —  l)d  = 

n 

.".  In  —  dn^  -\-  dn=2s  —  ln. 
.-.  dn'^-{2l  +  d)n==-2s. 
This  is  a  quadratic  with  n  for  the  unknown  number. 
Complete  the  square, 

4^2^2  _  0  +  (2Z  +  (^)2  =  (2^  +  df  -Sds. 
Extract  the  root, 

2dn-{2l  +  d)  =  ±  y/{2l  +  dY-Sds. 

2l  +  d±V{2l  +  df-8ds 

. .  n  = ^^ 

2d 

Note.  The  table  on  the  following  page  contains  the  results  of 
the  general  solution  of  all  possible  problems  in  arithmetical  series, 
in  which  three  of  the  numbers  a,  I,  d,  n,  s  are  given  and  two  required. 
The  student  is  advised  to  work  these  out,  both  for  the  results  obtained 
and  for  the  practice  gained  in  solving  literal  equations  in  which  the 
unknown  numbers  are  represented  by  letters  other  than  x,  y,  z. 


n 
2s  —  In 


298 


ALGEBRA. 


No. 

1 
2 
3 
4 

Given. 

Required. 

Results. 

a  dn 
ads 
a  n  s 
d  n  s 

I 

l  =  a  +  {n-V)d 

l=^-ld±^[2ds  +  {a-ldf]. 
l  =  '-^-a. 

^J     {n-l)d^ 
n            2 

5 

6 

7 
8 

a  dn 
adl 
a  n  I 
dnl 

s 

s  =  in[2a  +  {n-\)d]. 

l  +  a  ,  l^  —  a^ 
'         2      '     2d   ' 
s  =  (Z  +  a)| 

s  =  in[2l-{n-l)d]. 

9 
10 
11 
12 

dnl 
dn  s 
dls 
nls 

a 

a^l-{n-l)d. 

^_s      (n-l)d 

n           2 

a^^d±V{l  +  idY-2ds. 

a^^-^-l. 
n 

13 
14 
15 
16 

17 
18 
19 
20 

a  n  I 
a  n  s 
a  I  s 
nls 

d 

n-l 
^__2(s~an) 
n(n-l) 

^-'2s-l-a 
^__2(nl-s) 
n{n~l) 

adl 
ads 
a  I  s 
dls 

n 

n-.-^-%.l. 
d 

^      d-2a±V{2a-dY  +  Sds 
2d 

l  +  a 

^_2l  +  d±V{2l  +  df-8ds 
2d 

ARITHMETICAL    PROGRESSION.  299 

309.  The  arithmetical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
an  arithmetical  series. 

If  a  and  b  represent  two  numbers,  and  ^  their  arithmet- 
ical mean,  then,  by  the  definition  of  an  arithmetical  series, 
A  —a  =  b  —  A. 

310.  Sometimes  it  is  required  to  insert  several  arithmet- 
ical means  between  two  numbers. 

Insert  six  arithmetical  means  between  3  and  17. 

Here  the  whole  number  of  terms  is  eight;  3  is  the  first  term,  and 
17  the  eighth. 

By  I.,  11  =  3  + Id. 

d=2. 

The  series  is      3,     [5,     7,     9,     11,     13,     15,]     17, 
the  terms  in  brackets  being  the  means  required. 

311.  When  the  sum  of  a  number  of  terms  in  arithmet- 
ical progression  is  given,  it  is  convenient  to  represent : 

Three  terms  by         ^  —  y,     ^,     ^  +  y- 

Four  terms  by     x  —  St/,     x  —  y,     x-{-y,     x-^-Zy. 

The  sum  of  three  numbers  in  arithmetical  progression 
is  36,  and  the  square  of  the  mean  exceeds  the  product  of 
the  two  extremes  by  49.     Find  the  numbers. 

Let  X  —  y,  X,  X  +  y  represent  the  numbers. 
Then,  adding,  3  a;  =  36.       .-.  x  =  12. 

Putting  for  x  its  value,  the  numbers  are 

12 -y,     12,     12  +  2/. 
The  value  of  3/  is  ±  7  ;  and  the  numbers  are 

5,    12,    19;     or     19,    12,   5. 


300  ALGEBRA. 


Exercise  109. 

1.  Find  the  thirteenth  term  of  5,  9,  13 

ninth  term  of  —  3,  —  1,  1 

tenth  term  of  —  2,  —  5,  —  8 

eighth  term  of  a,  a  +  3 5,  a-\-6b 

fifteenth  term  of  1,  f ,  |- 

thirteenth  term  of  —  48,  —44,-40 

2.  The  first  term  of  an  arithmetical  series  is  3,  the  thir- 

teenth term  is  55.     Find  the  common  difference. 

3.  Find  the  arithmetical  mean  between:  (a.)  3  and  12; 

(b.)  -  5  and  17  ;  (c.)  o? -^  ah  -  b'  and  a' -  ab -{-  b\ 

4.  Insert  three  arithmetical  means  between  1  and  19;  and 

four  means  between  —  4  and  17. 

5.  The  first  term  of  a  series  is  2,  and  the  common  differ- 

ence \.     What  term  will  be  10  ? 

6.  The  seventh  term  of  a  series,  whose  common  difference 

is  3,  is  11.     Find  the  first  term. 

1,    Find  the  sum  of 

5  +  8  4- 11  + to  ten  terms. 

—  4  —  1  +  2  + to  seven  terms. 

a-l-4a  +  7a-j- tow  terms. 

f  +  tV  + 1*3-  + to  twenty-one  terms. 

1  +  2|-  -f  4-|-  + to  twenty  terms. 

8.  The  sum  of  six  numbers  of  an  arithmetical  series  is  27, 

and  the  first  term  is  1.     Determine  the  series. 

9.  How  many  terms  of  the  series  —  5  —  2  +1  + must 

be  taken  so  that  their  sum  may  be  63  ? 

10.    The  first  term  is  12,  and  the  sum  of  ten  terms  is  10. 
Find  the  last  term. 


SERIES.  301 

11.  The  arithmetical  mean  between  two  numbers  is  10, 

and  the  mean  between  the  double  of  the  first  and 
the  triple  of  the  second  is  27.     Find  the  numbers. 

12.  Find  the  middle  term  of  eleven  terms  whose  sum  is  66. 

13.  The  first  term  of  an  arithmetical  series  is  2,  the  com- 

mon difference  is  7,  and  the  last  term  79.  Find  the 
number  of  terms. 

14.  The  sum  of  fifteen  terms  of  an  arithmetical  series  is  600, 

and  the  common  difference  is  5.    Find  the  first  term. 

15.  Insert  ten  arithmetical  means  between  —  7  and  114. 

16.  The  sum  of  three  numbers  in  arithmetical  progression 

is  15,  and  the  sum  of  their  squares  is  83.     Find  the 

numbers. 

Let  x  —  y,  X,  X  -{-  y  represent  the  numbers. 

17.  Arithmetical  means  are  inserted  between  5  and  23,  so 

that  the  sum  of  the  first  two  is  to  the  sum  of  the  last 
two  as  2  is  to  5.     How  many  means  are  inserted  ? 

18.  Find  three  numbers  of  an  arithmetical  series  whose 

sum  shall  be  21,  and  the  sum  of  the  first  and  second 
shall  be  |  of  the  sum  of  the  second  and  third. 

19.  Find  three  numbers  whose  common  difi'erence  is  1, 

such  that  the  product  of  the  second  and  third  ex- 
ceeds that  of  the  first  and  second  by  -^. 

20.  How  many  terms  of  the  series  1,  4,  7 must  be 

taken,  in  order  that  the  sum  of  the  first  half  may 
bear  to  the  sum  of  the  second  half  the  ratio  10  :  31  ? 

21.  A  travels  uniformly  20  miles  a  day;  B  starts  three 

days  later,  and  travels  8  miles  the  first  day,  12  the 
second,  and  so  on,  in  arithmetical  progression.  In 
how  many  days  will  B  overtake  A  ? 


302  ALCxEBEA. 

22.  A  number  consists  of  three  digits  which  are  in  arith- 

metical progression ;  and  this  number  divided  by 
the  sum  of  its  digits  is  equal  to  26 ;  but  if  198  be 
added  to  it,  the  digits  in  the  units'  and  hundreds' 
places  will  be  interchanged.     Required  the  number. 

23.  The  sum  of  the  squares  of  the  extremes  of  four  numbers 

in  arithmetical  progression  is  200,  and  the  sum  of  the 
squares  of  the  means  is  136.  What  are  the  numbers? 

24.  Show  that  if  any  even  number  of  terms  of  the  series  1, 

3,  5 be  taken,  the  sum  of  the  first  half  is  to  the 

sum  of  the  second  half  in  the  ratio  1  :  3. 

25.  A  and  B  set  out  at  the  same  time  to  meet  each  other 

from  two  places  343  miles  apart.  Their  daily  jour- 
neys are  in  arithmetical  progression,  A's  increase 
being  2  miles  each  day,  and  B's  decrease  being  5 
miles  each  day.  On  the  day  at  the  end  of  which 
they  met,  each  travelled  exactly  20  miles.  Find  the 
duration  of  the  journey. 

26.  Suppose  that  a  body  falls  through  a  space  of  IQ^  feet 

in  the  first  second  of  its  fall,  and  in  each  succeeding 
second  32-|-  more  than  in  the  next  preceding  one. 
How  far  will  a  body  fall  in  20  seconds? 

27.  The  sum  of  five  numbers  in  arithmetical  progression 

is  45,  and  the  product  of  the  first  and  fifth  is  f  of 
the  product  of  the  second  and  fourth.  Fird  the 
numbers. 

28.  If  a  full  car  descending  an  incline  draw  up  an  empty 

one  at  the  rate  of  1^  feet  the  first  second,  4|-  feet  the 
next  second,  7^  feet  the  third,  and  so  on,  how  long 
will  it  take  to  descend  an  incline  150  feet  in  length? 
What  part  of  the  distance  will  the  car  have  de- 
scended in  the  first  half  of  the  time  ? 


geometrical  progression.  303 

Geometrical  Progression. 

312.  A  series  is  called  a  geometrical  series  or  a  geometrical 
progression  when  each  succeeding  term  is  obtained  by  mul- 
tiplying the  preceding  term  by  a  constant  multiplier. 

The  general  representative  of  such  a  series  will  be 


a,  ar,  ar"^,  ar^,  ar* 


in  which  a  is  the  first  term  and  r  the  constant  multiplier 
or  ratio. 

The  terms  increase  or  decrease  in  numerical  magnitude 
according  as  r  is  numerically  greater  than  or  numerically 
less  than  unity. 

313.  The  nth  Term.  Since  the  exponent  of  ?•  increases 
by  one  for  each  succeeding  term  after  the  first,  the  expo- 
nent will  always  be  one  less  than  the  number  of  the  term, 
so  that  the  nth  term  is  ar"^'^. 

If  the  nth  term  is  represented  by  /,  we  have 

I  =  ar^'-K  I. 

314.  Sum  of  the  Series.  If  I  represents  the  nth  term,  a  the 
first  term,  n  the  number  of  terms,  r  the  common  ratio,  and 
s  the  sum  of  n  terms,  then 

s  =  a -\- ar  -\-  ar^  -\- ar'^~^. 

Multiply  by  r, 

rs  =  ar-\-  ar^  +  ar^  -f ar""'  -f  ar"". 

Subtracting  the  first  equation  from  the  second, 
rs~  s  =  ar^  —  a, 
or  (r  —  1)  s  =  a  (r"  —  1). 

,.  ,^«fr"-l).  IT. 

7—1 


304  ALGEBRA. 

Since  a?'""^  =  I,  it  follows  that  ar^  —  rl,  and  11.  may  be 
written 

s  = — •  III. 

r  —  1 


315.  From  the  two  equations  I.  and  IL,  or  the  two 
equations  I.  and  III.,  any  two  of  the  five  numbers  a,  r,  I, 
n,  s,  may  be  found  when  the  other  three  are  given. 

(1)  The  first  term  of  a  geometrical  series  is  3,  the  last 
term  192,  and  the  sum  of  the  series  381.  Find  the  num- 
ber of  terms  and  the  ratio. 


From  I.  and  III.,               192  =  3  r»-", 

(1) 

381      192r-3 
r-  1 

(2) 

From  (2),                                r  =  2. 

Substituting  in  (1),           2"-'  =  64  =  2". 

.-.  n  =  7. 

The  series  is    3,     6,     12,     24,    48,    96,     192. 

(2)  Find  I  when  r,  n,  s  are  given. 

From  I.,                                  a  =  ——■ 

"'-»'. 

(-i)-^^;:/^'- 

.    ,(r-l)r'-h 

r»-  1 

Note.  The  table  on  page  305  contains  the  results  of  all  possible 
problems  in  geometrical  series  in  which  three  of  the  numbers  a,  r,  I, 
n,  s,  are  given  and  the  other  two  required,  with  the  exception  of 
those  in  which  n  is  required ;  these  last  require  the  use  of  logarithms 
with  which  the  student  is  supposed  to  be  not  yet  acquainted. 


GEOMETRICAL   PROGRESSION. 


305 


No. 

Given. 

Required. 

Results. 

1 
2 
3 
4 

a  r  n 
a  r  s 
a  n  s 
r  n  s 

I 

I  =  ar'^-K 
^_a  +  (r-l)s 

r 
Z(s-^)«-i-a(s-a)«-i  =  0. 
;_(r-l)sr"  1 
r"-  1 

5 
6 

7 
8 

a  r  n 
arl 
a  n  I 
r  nl 

s 

r-1 
r-1 

?■»  -  r"  -' 

9 
10 

11 
12 

ml 
r  n  s 

r  I  s 
nl  s 

a 

a     '^1^^^. 
r«-  1 

a  =  W-(r-l)s. 

13 

14 
15 
16 

an  I 
a  n  s 
als 
n  I  s 

r 

'a 
a           a 

r"-_^r"-»+-^  =  0. 
s-l             s-  I 

316.  The  geometrical  mean  between  two  numbers  is  the 
number  which  stands  between  them,  and  makes  with  them 
a  geometrical  series. 


306  ALGEBRA. 

If  a  and  b  denote  two  numbers,  and  G^  their  geometrical 
mean,  then,  by  the  definition  of  a  geometrical  series, 

a      G 

317.  Sometimes  it  is  required  to  insert  several  geometri- 
cal means  between  two  numbers. 

Insert  three  geometrical  means  between  3  and  48. 
Here  the  whole  number  of  terms  is  five  ;  3  is  the  first  term,  and  48 
the  fifth.  -^ 

By  I.,  48  =  3r*, 

r*  =  16, 
r  =  ±  2. 
The  series  is  one  of  the  following  : 

3,     [     6,     12,         24,]    48; 

3,     [-6,     12,     -24,]    48. 

The  terms  in  brackets  are  the  means  reqijired. 

318.  Infinite  Geometrical  Series.  When  r  is  less  than  1, 
the  successive  terms  become  numerically  smaller  and 
smaller ;  by  taking  n  large  enough  we  can  make  the  nth. 
term,  ar**~^,  as  small  as  we  please,  although  we  cannot 
make  it  absolutely  zero. 

The  sum  of  n  terms, ,  may  be  written ; 

r  —  1  l~rl— r 

this  sum  differs  from by  the  fraction ;  by  taking 

l-r    ^  1-r      ^  ^ 

enough  terms  we  can  make  /,  and  consequently  this  dif- 
ference, as  small  as  we  please ;  the  greater  the  number  of 

terms  taken,  the  nearer  does  their  sum   approach 

1  —  r 

Hence  ^  is  called  the  sum  of  an  infinite  number  of 

1  —  r 

terms  of  the  series. 


GEOMETRICAL    PROGRESSION.  307 

(1)  Find  the  sum  of  the  infinite  series 

1-1  +  1-1+ 

Here,  a  =  1,  r  = 

2 

1  2 

The  sum  of  the  series  is  or  — 

1+J       3 

We  find  for  the  sum  of  n  terms    o  ~  o  (  ~  9  )       >  ^"^^  ^^^  ^^^' 
dently  approaches  -  as  n  is  increased. 

(2)  Find  the  value  ofthe  recurring  decimal  0.12135135 


Consider    first    the    part    that    recurs ;     this    may    be    written 

135 
135     ^ 135 _^ ^  ^^^  ^^^  g^^  ^f  ^j^jg  gg^.gg  -g    100000 


100000      100000000  J 1_ 

^  3000 

which  reduces  to .    Adding  0.12,  the  part  that  does  not  recur,  we 

•  449 

obtain  for  the  value  of  the  decimal  -— — . 

3700 


Exercise  110. 

1.  Find  the  seventh  term  of  2,  6,  18 

sixth  term  of  3,  6,  12 

ninth  term  of  6,  3,  1^ 

eighth  term  of  1,  —2,  4 

twelfth  term  of  x^,  x\,  x" 

fifth  term  of  4a,  —  6wa^  9mV 

2.  Find  the  geometrical  mean  between  18^'y  and  300^2. 

3.  Find  the  ratio  when  the  first  and  third  terms  are  5 

and  80  respectively. 

4.  Insert  two  geometrical  means  between  8  and  125  ;  and 

three  between  14  and  224. 


308  ALGEBRA. 

5.  U  a  =  2  and  r  =  S,  which  term  will  be  equal  to  162  ? 

6.  The  fifth  term  of  a  geometrical  series  is  48,  and  the 

ratio  2.     Find  the  first  and  seventh  terms. 

7.  Find  the  sum  of 

3  +  6  +  12  + to  eight  terms. 

1  —  3  +  9  — to  seven  terms. 

8  +  4  +  2  + to  ten  terms. 

0.1  +  0.5  +  2.5  + to  seven  terms. 

m  —  ^  +  ~  — to  five  terms. 

4      16 

8.  The  population  of  a  city  increases  in  four  years  from 

10,000  to  14,641.     What  is  the  rate  of  increase  ? 

9.  The  sum  of  four  numbers  in  geometrical  progression  is 

200,  and  the  first  term  is  5.     Find  the  ratio. 

10.  Find  the  sum  of  eight  terms  of  a  series  whose  last  term 

is  1,  and  fifth  term  ^. 

11.  In  an  odd  number  of  terms,  show  that  the  product  of 

the  first  and  last  will  be  equal  to  the  square  of  the 
middle  term. 

12.  The  product  of  four  terms  of  a  geometrical  series  is  4, 

and  the  fourth  term  is  4.     Determine  the  series. 

13.  If  from  a  line  one-third  be  cut  off",  then  one-third  of 

the  remainder,  and  so  on,  what  fraction  of  the  whole 
will  remain  when  this  has  been  done  five  times  ? 

14.  Of  three  numbers  in  geometrical  progression,  the  sum 

of  the  first  and  second  exceeds  the  third  by  3,  and 
the  sum  of  the  first  and  third  exceeds  the  second  by 
21.     What  are  the  numbers  ? 

15.  Find  two  numbers  whose  sum  is  3i  and  geometrical 

mean  1^. 


GEOMETRICAL    PROGRESSION.  309 

16.  A  glass  of  wine  is  taken  from  a  decanter  that  holds  ten 

glasses,  and  a  glass  of  water  poured  in.  After  this 
is  done  five  times,  what  part  of  the  contents  is  wine  ? 

17.  There  are  four  numbers  such  that  the  sum  of  the  first 

and  the  last  is  11,  and  the  sum  of  the  others  is  10. 
The  first  three  of  these  four  numbers  are  in  arith- 
metical progression,  and  the  last  three  are  in  geomet- 
rical progression.     Find  the  numbers. 

18.  Find  three  numbers  in  geometrical  progression  such 

that  their  sum  is  13  and  the  sum  of  their  squares 
is  91. 

19.  The  difiference  between  two  numbers  is  48,  and  the 

arithmetical  mean  exceeds  the  geometrical  by  18. 
Find  the  numbers. 

20.  There  are  four  numbers  in  geometrical  progression, 

the  second  of  which  is  less  than  the  fourth  by  24, 
and  the  sum  of  the  extremes  is  to  the  sum  of  the 
means  as  7  to  3.     Find  the  numbers. 

21.  A  number  consists  of  three  digits  in  geometrical  pro- 

gression. The  sum  of  the  digits  is  13 ;  and  if  792 
be  added  to  the  number,  the  digits  in  the  units'  and 
hundreds'  places  will  be  interchanged.  Find  the 
number. 

22.  Find  the  sum  of  each  of  the  infinite  series  : 


4+2+    1  + 

2-li  +  l- 

i+  i+  i  + 

0.1  +  0.01  +  0.001  + 

i-l'T  +  TV- 

0.868686 

1-  *  +  A- 

0,54444 

i  +  ^  +  is  + 

0.83636 

310  ALGEBRA. 

Haemonical  Progression. 

319.  A  series  is  called  a  harmonical  series,  or  a  harmonical 
progression,  when  the  reciprocals  of  its  terms  form  an  arith- 
metical series. 

The  general  representative  of  such  a  series  will  be 

1^    _J_^         1         ^  1 

a     a-\-d    a-\-2d  a-\-{n  —  l)d 

Questions  relating  to  harmonical  series  are  best  solved 
by  writing  the  reciprocals  of  its  terms,  and  thus  forming  an 
arithmetical  series. 

320.  If  a  and  h  denote  two  numbers,  and  H  their  har- 
monical mean,  then,  by  the  definition  of  a  harmonical  series, 

J^_l_l_  J^ 
H     a     b      H 

.     2  _1  .  l_a  +  b 
"  H     a     b        ab 


321.  Sometimes  it  is  required  to  insert  several  harmoni- 
cal means  between  two  numbers. 

Insert  three  harmonical  means  between  3  and  18. 

Find  the  three  arithmetical  means  between  -  and  — ■ 

19  14    9  • 

These  are  found  to  be  — ,  — ,  — ;  therefore,  the  harmonical  means 

72  72  72 
72   72   72 

19'  14    9 

322.  Since,  §§  309,  316,  320, 

2  a  +  b 

we  have  A  \  O  ^  O  \  H. 


HARMONICAL    PROGRESSION.  311 

Exercise  111. 

1.  Insert  four  harmonical  means  between  2  and  12. 

2.  Find  two  numbers  whose  difference  is  8  and  the  har- 

monical mean  between  them  1|-. 

3.  Find  the  seventh  term  of  the  harmonical  series  3,  3^, 

4 

4.  Continue  to  two  terms  each  way  the  harmonical  series, 

two  consecutive  terms  of  which  are  15,  16. 

5.  The  first  two  terms  of  a  harmonical  series  are  5  and  6. 

Which  term  will  equal  30  ? 

6.  The  fifth  and  ninth  terms  of  a  harmonical  series  are  8 

and  12.     Find  the  first  four  terms. 

7.  The  difference  between  the  arithmetical  and  harmonical 

means  between  two  numbers  is  1-|,  and  one  of  the 
numbers  is  four  times  the  other.    Find  the  numbers. 

8.  Find   the  arithmetical,  geometrical,  and  harmonical 

means  between  two  numbers  a  and  h ;  and  show  that 
the  geometrical  mean  is  a  mean  proportional  between 
the  arithmetical  and  harmonical  means.  Also,  ar- 
range these  means  in  order  of  magnitude. 

9.  The  arithmetical  mean  between  two  numbers  exceeds 

the  geometrical  by  13,  and  the  geometrical  exceeds 
the  harmonical  by  12.     What  are  the  numbers? 

10.  The  sum  of  three  terms  of  a  harmonical  series  is  11,  and 

the  sum  of  their  squares  is  49.     Find  the  numbers. 

11.  .When  a,  5,  c  are  in  harmonical  progression,  show  that 

a\c\\a  —  h\h~c. 


CHAPTER  XXIV. 
INDETERMINATE   COEFFICIENTS. 

323.  Convergent  and  Divergent  Series.     By  performing  the 

indicated  division,  we  obtain  from  the  fraction  :; the 

1  —x 

infinite  series  1 -\- x -}- x^  -{- x^  -{- This  series,  however, 

is  not  equal  to  the  fraction  for  all  values  of  x. 

324.  If  X  is  numerically  less  than  1,  the  series  is  equal  to 
the  fraction.  In  this  case  we  can  obtain  an  approximate 
value  for  the  sum  of  the  series  by  taking  the  sum  of  a  num- 
ber of  terms ;  the  greater  the  number  of  terms  taken,  the 
nearer  will  this  approximate  sum  approach  the  value  of  the 
fraction.  The  approximate  sum  will  never  be  exactly  equal 
to  the  fraction,  however  great  the  number  of  terms  taken  ; 
but  by  taking  enough  terms,  it  can  be  made  to  differ  from 
the  fraction  as  little  as  we  please. 

Thus,  if  X  —  ^,  the  value  of  the  fraction  is  2,  and  the 
series  is  i      i      i 

The  sum  of  four  terms  of  this  series  is  IJ ;  the  sum  of 
five  terms,  l-J-f- ;  the  sum  of  six  terms,  1|-|- ;  and  so  on. 
The  successive  approximate  sums  approach,  biit  never 
reach,  the  finite  value  2. 

325.  An  infinite  series  is  said  to  be  convergent  when  the 
sum  of  the  terms,  as  the  number  of  terms  is  indefinitely  in- 
creased, approaches  some  fixed  finite  value;  this  finite  value 
is  called  the  sum  of  the  series. 


INDETERMINATE   COEFFICIENTS.  313 

326.  In    the    series    l-j-x-^x'^-^-cc^-j- suppose   x 

numerically  greater  than  1.  In  this  case,  the  greater  the 
number  of  terms  taken,  the  greater  will  their  sum  be;  by- 
taking  enough  terms,  we  can  make  their  sum  as  large  as 
we  please.  The  fraction,  on  the  other  hand,  has  a  definite 
value.  Hence,  when  x  is  numerically  greater  than  1,  the 
series  is  not  equal  to  the  fraction. 

Thus,  if  a;  =  2,  the  value  of  the  fraction  is  —  1,  and  the 
series  is 

1  +  2  +  4  +  8  + 

The  greater  the  number  of  terms  taken,  the  larger  the  sum. 
Evidently  the  fraction  and  the  series  are  not  equal. 

327.  In  the  same  series  suppose  x=l.     In  this  case  the 

fraction  is ^  —  j\'    ^^^  ^^^  series  1  +  1  +  1  +  1  + 

The  more  terms  we  take,  the  greater  will  the  sum  of  the 
series  be,  and  the  sum  of  the  series  does  not  approach  a 
fixed  finite  value. 

If  X,  however,  is  not  exactly  1,  but  is  a  little  less  than  1, 

the  value  of  the  fraction will  be  very  great,  and  the 

fraction  will  be  equal  to  the  series. 

Suppose  x  =  —  I.    In  this  case  the  fraction  is =  -, 

1  +  12 

and  the  series  1  —  1  +  1  —  1  + If  we  take  an  even 

number  of  terms,  their  sum  is  0 ;  if  an  odd  number,  their 

sum  is  1.     Hence  the  fraction  is  not  equal  to  the  series. 

328.  A  series  is  said  to  be  divergent  when  the  sum  of  the 
terms,  as  the  number  of  terms  is  indefinitely  increased, 
either  increases  without  end,  or  oscillates  in  value  without 
approaching  any  fixed  finite  value. 


314  ALGEBRA. 

No  reasoning  can  be  based  on  a  divergent  series ;  hence, 
in  using  an  infinite  series  it  is  necessary  to  make  such 
restrictions  as  will  cause  the  series  to  be  convergent.    Thus, 

we  can  use  the  infinite  series  1 -\- x -}- x^ -{- x^ -{- when, 

and  only  w^hen,  x  lies  between  +  1  and  —  1. 

329.  A  series,  ax  +  hx"^  +  cx^  +  <^^*  + ,  in  which  the  co- 
efficients a,  b,  c,  d are  finite,  may,  by  taking  x  sufficiently 

small,  be  made  less  than  any  assigned  value. 

For  if  q  is  any  assigned  value,  and  k  the  greatest  of  the  coeffi- 
cients a,  6,  c,  ,  then 

ax  +  hx^  +  ca^  + <  ^x  +  hx^  +  hx^  + 

But  kx^hx^  +  kx^^ =._^ 

1  — a; 

(as  is  evident  by  dividing  kxhy  \  —  x). 

:.  ax  +  hx"^  +  cx^  + <  — —,  if  x  is  taken  less  than  1. 

1—x 

Hence,  if  — —  be  taken  less  than  q, 

1  —  x 

that  is,  if  re  <     ^   ■> 

q^-k 

then  ax  +  hx"^  -\-  cx^  + will  be  less  than  q. 


330.  Theorem  of  Indeterminate  OoefiBcients.  If  two  series, 
arranged  hy  powers  of  x,  are  equal  for  all  values  of  x  that 
mahe  both  series  convergent,  the  corresponding  coefficients  are 
equal  each  to  each. 

For,  if  ^  +  ^a;  -I-  Cx^  + =  A^  +  B^x  +  (7V  + , 

by  transposition, 

A-A^^{B'-B)x^  {C  -  C) x2  + 

Now  by  taking  x  sufficiently  small,  the  right  side  of  this  equation 
can  be  made  less  than  any  assigned  value  whatever,  and  therefore 
less  than  A  —  A^,  \i  A  —  A^  has  any  value  whatever.  Hence  A  —  A^ 
cannot  have  any  value. 


INDETERMINATE    COEFFICIENTS.  315 

.-.  A-A'  =  0  or  A  =  A'. 

Hence,       Bx  +  Cx^  +  Da^  + =  B^x  +  C^x^  +  DV  + 

or  {B  -B')x  =  {(y-  C)x^  +  {D'  -  D)x^  + ; 

by  dividing  by  x, 

B-B'  =  {C'-0)x  +  {D'  -D)x^  + ; 

and,  by  the  same  proof  as  for  A  —  A^, 
B-B'^0  or  B  =  B^. 
In  like  manner, 

C=C\  D  =  D\  and  so  on. 

Hence,  the  equation 

A -{- Bx -{- Cx" -\- =  ^'  4-  B'x  +  (7V  + , 

if  true  for  all  finite  values  of  x,  is  an  identical  equation; 
that  is,  the  coefficients  of  like  powers  of  x  are  the  same. 

Expand  - — '     ^  ^  in  ascending  powers  of  x. 

Assume         ^  +  ^^    =  A  ^  Bx  ^- Cx"  ^-  Dx^  + ; 

l-\-x-\-x^ 

then,  by  clearing  of  fractions, 

2  +  Zx  =  A  +  Bx  -v  Cx^  -V  Da?  + 

■\-Ax  +  Bx^  +  Ca^  + 

+  Ax''  +  Bx^  + 

.:  2  +  3x  =  A  +  {B  +  A)x  +  {C  +  B  +  A)x^  +  {D  +  C  +  B)a?  + 

.-.  A  =  2,  ^  +  ^  =  3,  C+^  +  ^  =  0,  i)  +  C  +  5  =  0; 
whence  B=\,  C=  — 3,  D  =  2\  and  so  on. 

.-.  _2±l^  =  2  +  a;-3x2  +  2x3  + 

1  +  X  +  cc^ 

The  series  is  of  course  equal  to  the  fraction  for  only  such  values 
of  x  as  make  the  series  convergent. 


316  ALGEBRA. 

Note.  In  employing  the  method  of  Indeterminate  Coefficients, 
the  form  of  the  given  expression  must  determine  what  powers  of  the 
variable  x  must  be  assumed.  It  is  necessary  and  sufficient  that  the 
assumed  equation,  when  simplified,  shall  have  in  the  right  member 
all  the  powers  of  x  that  are  found  in  the  left  member. 

If  any  powers  of  x  occur  in  the  right  member  that  are  not  in  the 
left  member,  the  coefficients  of  these  powers  in  the  right  member  will 
vanish,  so  that  in  this  case  the  method  still  applies  ;  but  if  any  powers 
of  X  occur  in  the  left  member  that  are  not  in  the  right  member,  then 
the  coefficients  of  these  powers  of  x  must  be  put  equal  to  0  in  equating 
the  coefficients  of  like  powers  of  x ;  and  this  leads  to  absurd  results. 
Thus,  if  it  were  assumed  that 

^'^^^    =  Ax  +  Bx^  +  Ca^  + , 

]  +x  +  x^ 

there  would  be  in  the  simplified  equation  no  term  on  the  right  cor- 
responding to  2  on  the  left ;  so  that,  in  equating  the  coefficients  of 
like  powers  of  a;,  2,  which  is  2x®,  would  have  to  be  put  equal  to  Ox"; 
that  is,  2  =  0,  an  absurdity.  • 

Exercise  112. 
Expand  to  four  terms  in  ascending  powers  of  x : 

1.         ^  2.    -1+^.      3.    ^~2^       -  ^~" 


5. 


2x  +  Sx''  l  +  Sx-x'  l-2a;  +  3r^ 


331.  Partial  Practions.  To  resolve  a  fraction  into  partial 
fractions  is  to  express  it  as  the  sum  of  a  number  of  frac- 
tions of  which  the  respective  denominators  are  the  factors 
of  the  denominator  of  the  given  fraction.  This  process  is 
the  reverse  of  the  process  of  adding  fractions  which  have 
different  denominators. 

Resolution  into  partial  fractions  may  be  easily  accom- 
plished by  the  use  of  indeterminate  coefficients. 

In  decomposing  a  given  fraction  into  its  simplest  partial 


INDETERMINATE   COEFFICIENTS.  317 

fractions,  it  is  important  to  determine  what  form  the  assumed 

fractions  must  have. 

Since  the  given  fraction  is  the  sum  of  the  required  par- 
tial fractions,  each  assumed  denominator  must  be  a  factor 
of  the  given  denominator ;  moreover,  all  the  factors  of  the 
given  denominator  must  be  taken  as  denominators  of  the 
assumed  fractions. 

Since  the  required  partial  fractions  are  to  be  in  their 
simplest  form  incapable  of  further  decomposition,  the  nu- 
merator of  each  required  fraction  must  be  assumed  with 
reference  to  this  condition.  Thus,  if  the  denominator  is 
.r"  or  (x  ±  ay,  the  assumed  fraction  must  be  of  the  form 

—  or ;  tor  it  it  had  the  lorm ■ or ■ , 

a;"       {x  zb  aY  x""  (x  ±  ay 

it  could  be  decomposed  into  two  fractions,  and  the  partial 

fractions  would  not  be  in  the  simplest  form  possible. 

When  all  the  monomial  factors,  and  all  the  binomial 

factors,  of  the  form  x  ±  a  have  been  removed  from  the 

denominator  of  the  given  expression,  there  may  remain 

quadratic  factors  which  cannot  be  further  resolved ;  and 

the  numerators  corresponding  to  these  quadratic  factors 

may  each  contain  the  first  power  of  x,  so  that  the  assumed 

fractions  must  have  either  the  form  — — -,  or  the 

formi£+^.  ^  "''"  +  ' 

x'-i-b 

3       . 

(1)  Resolve  — into  partial  fractions. 

x^  -{-1 

Since  a^  +  l  =  {x  +  l){x^  —  a;  +  1),  the  denominators  will  be  a;  +  1 
and  rc^  —  a;  +  1. 

Assume  -- — -  = +    ^  "^  ^ — -  ; 

ar  +  1     X  +  1     x^  —  X  +  1 

then  3  =  A{x''-x  +  l)  +  {Bx  +  C){x  +  1) 

=  {A  +  B)x^  +  {B  +  C-  A)x  +  {A  +  C); 


318  ALGEBRA. 

whence,  A  +  C^  3,  B  +  C -  A  =  0,  A  +  B  =  0, 

and  A^l,  B  =  -l,  (7-2. 

3  1  x-2 


Therefore. 


a;'  +  l      X  +  1     x"^  —  X  +  1 


(2)   Resolve — — into  partial  fractions. 

x^  (x  +  l)'^ 

The  denominators  may  be  x,  x"^,  a;  +  1,  (a;  +  l)^. 


x''{x  +  lf  X      x""     x  +  1     {x  +  iy 

.:  4:a^-x'^-3x-2  =  Ax{x  +  If  +  B{x  +  l)'^  +  Cx''{x  +  1)  +  Dx^ 

=  {A  +  C)x3  -f  (2^  +  ^  +  C+  D)x'  -\-{A-v2B)x  +  B- 

whence,  J.  +  C  =-  4, 

2^  +  J5+C+Z>--l, 

^  +  25  =  -3, 

B  =  -2- 

or  5  -  -  2,  ^  =  1,  C=  3,  i>  -  -  4. 

Therefore,    4.^-.^-3.-2^1_2^ i_, 

a;2(a;  +  l)^  a;     a;^      a;  +  1      {x  +  1)^ 

Exercise  113. 
Resolve  into  partial  fractions  : 

1x^-1  ^  x~2  ^      3:^2-4 


x"- 

■2>x- 

-10 

3 

x'- 

~l 

x"- 

■  X  — 

■3 

(a;+4)(a;-5)  a;^-3a;-10  x\x -}- b) 

2.    . ?— ^.      5.    ^1^.  8.  ^^'- 


(x+S)(x-{-4:)  a;'-l  *    {x-l)Xx+2) 

3  537-1  g     g;'  -  a:  -  3         ^     2a;^-7a;  +  l 

(2^-l)(a;-5)'      *     a;(a;^-4)'  *  x'-l 


CHAPTER  XXV. 
BINOMIAL  THEOREM. 

332.  Binomial  Theorem,  Positive  Integral  Exponent.  By  suc- 
cessive multiplications  we  obtain  the  following  identities : 

{a  +  hf  =  a'  4-  ^o^h  +  Zah"  +  ¥  ; 

{a  +  hy  =  a'-\-  Wh  +  6  a'h''  +  4  a^H  h\ 

The  expressions  on  the  right  may  be  written  in  a  form 
better  adapted  to  show  the  law  of  their  formation : 

(a  +  iy  =  a^  +  2a5+|;|.J'; 

ia  +  bY:^a^  +  ia?b  +  ^^aV+^^^ab'  +  ^W:^b\ 

Note.  The  dot  between  the  Arabic  figures  means  the  same  as  the 
sign  X. 

333.  Let  n  represent  the  exponent  of  {a  +  h)  in  any  one 
of  these  identities ;  then,  in  the  expressions  on  the  right, 
we  observe  that  the  following  laws  hold  true : 

I.  The  number  of  terms  is  7i+  1. 

II.  The  first  term  is  a",  and  the  exponent  of  a  is  one 
less  in  each  succeeding  term. 

The  first  power  of  h  occure  in  the  second  term,  the 
second  power  in  the  third  term,  and  the  exponent  of  b  is 
one  greater  in  each  succeeding  term. 

The  sum  of  the  exponents  of  a  and  h  in  any  term  is  n. 


320  ALGEBRA. 

III.  The  coefficient  of  the  first  term  is  1 ;  of  the  second 
term,  n ;  of  the  third  term,  '^\~    ^  ;  and  so  on. 

334.  Consider  the  coefficient  of  any  term ;  the  number 
of  factors  in  the  numerator  is  the  same  as  the  number  of 
factors  in  the  denominator,  and  the  number  of  factors  in 
each  is  the  same  as  the  exponent  of  b  in  that  term ;  this 
exponent  is  one  less  than  the  number  of  the  term. 

335.  Proof  of  the  Theorem.  To  show  that  the  laws  of  §  333 
hold  true  when  the  exponent  is  any  positive  integer  : 

We  know  that   the   laws   hold  for  the  fourth  power; 
suppose,  for  the  moment,  that  they  hold  for  the  ki\i  power. 
We  shall  then  have 

{a  +  hf  =  a^  +  M-'h  +  Mlzil)  ^^-252 

_^^(y(;-l)(^-2)^fe_3^3_^ (1) 

Multiply  both  members  of  (1)  by  a  +  J ;  the  result  is 
^  ^ (a  +  J)*+'  =  a^+'  +  (^  +  1) a'h  +  (l+i)i^ oJ^-^b' 

-|_  (^+1)^(^-   -1)  ^^-2^3     ,     /2) 

l-2'3  ^  ^  ^ 

In  (1)  put  ^  -}-  1  for  /; ;  this  gives 

(a  +  5)^+^  =  a^+^  +  (>^  4-  1)  o!^b  +  (l+IX^  +  1  -  1)  ^*-i^2 

X '  A 

(^  +  l)(^  +  l-l)(/^  +  l-2)   ,_„3  , 

^         r2^3         "  *"^ 

=  a'+'  +  (yt  +  1)  a'5  +  (ijill^  ^j-ij2 
1  '  A 

1  •  2- 3  ^  ^ 

Equation  (3)  is  seen  to  be  the  same  as  equation  (2). 


BINOMIAL    THEOREM.  321 

Hence  (1)  holds  when  we  put  Jc -^  1  ior  k;  that  is,  if  the 
laws  of  §  833  hold  for  the  ^th  power,  they  must  hold  for 
the  {k  -f  l)th  power. 

But  the  laws  hold  for  the  fourth  power ;  therefore  they 
must  hold  for  the  fifth  power. 

Holding  for  the  fifth  power,  they  must  hold  for  the  sixth 
power ;  and  so  on  for  any  positive  integral  power. 

Therefore  they  must  hold  for  the  nth.  power,  if  n  is  a 
positive  integer ;  and  we  have 

(a  +  bf  =  a"  +  na^-'b  ^^(^  "  l)^n-2^2 

^n(n-l)(n-2)^„_,^,_^ ^^^ 

1  '  A  '  o 

Note.  The  above  proof  is  an  example  of  a  proof  by  mathematical 
induction. 


336.  This  formula  is  known  as  the  binomial  theorem. 

The  expression  on  the  right  is  known  as  the  expansion  of 
(a  -\-  by ;  this  expansion  is  a,  finite  series  when  w  is  a  positive 
integer.     That  the  series  is  finite  may  be  seen  as  follows : 

In  writing  out  the  successive  coefficients  we  shall  finally 
arrive  at  a  coefficient  which  contains  the  factor  n  —  n;  the 
corresponding  term  will  vanish.  The  coefficients  of  all  the 
succeeding  terms  likewise  contain  the  factor  n  —  n,  and 
therefore  all  these  terms  will  vanish. 

337.  If  a  and  b  are  interchanged,  the  identity  (A)  may 
be  written 

(a  +  by  - (5  +  af  =^b^  +  nb^-'a  +  ^^^~^^ b'^'W 

7i(n-l)(n-_2)^„_3  3  ,  

^  1-2-3  ^ 


322  ALGEBRA. 

This  last  expansion  is  the  expansion  of  (A)  written  in 
reverse  order.  Comparing  the  two  expansions,  we  see 
that:  the  coefficient  of  the  last  term  is  the  same  as  the 
coefficient  of  the  first  term ;  the  coefficient  of  the  last  term 
but  one  is  the  same  as  the  coefficient  of  the  first  term  but 
one ;  and  so  on. 

In  general,  the  coefficient  of  the  rth  term  from  the  end 
is  the  same  as  the  coefficient  of  the  rth  term  from  the 
beginning.  In  writing  out  an  expansion  by  the  binomial 
theorem,  after  arriving  at  the  middle  term,  we  can  shorten 
the  work  by  observing  that  the  remaining  coefficients  are 
those  already  found,  taken  in  reverse  order. 

338.  If  h  is  negative,  the  terms  which  involve  even 
powers  of  b  will  be  positive,  and  those  which  involve  odd 
powers  of  h  negative.     Hence, 

(a  -  by  =  a"  -  naJ'-'b  -f  ^i^?— i)  a^-'^h' 

_n(n-l){n-2)^,_,^,^_        (B) 
s. '  A  '  o 

Also,  putting  1  for  a  and  x  for  5,  in  (A)  and  (B), 
(1  +  ^)"=!+^^  +  ^^!'"^^^' 

^n(.-lXn^-2)^3^ (0) 


{l  —  xy  =  \  —  nx-\- 


n(n  —  1) 
1-2 


_n(n-l)(n-2)^      j^ 

1-2-3  ^  ^^ 


BINOMIAL   THEOREM.  323 

339.   Examples. 

(1)  Expand  (1  +  2xf. 

In  (0)  put  2  a;  for  a;  and  5  for  n.     The  result  is 
(1  +  2xf  =  1  +  5(2x)  +  1^4x2  +  1^  Sa;' 

1-2-3-4  1-2-3-4-5 

=  1  +  10a;  +  40a^2  ^  gOa;^  ^  gOa;*  +  320^. 

(2)  Expand  to  three  terms  [ ]  • 

Put  a  for  1  and  i  for  —  ;  then,  by  (B), 
X  3 

(a  -hf  =  a^-Q  a^h  +  Iba'h''  +  ...- 
Replacing  a  and  h  by  their  values, 

=  i  _  1  +  20  _ 


340.  Any  Required  Term.  From  (A)  it  is  evident  (§  335) 
that  the  (r  +  l)tli  term  of  the  expansion  of  {a  +  6)"  is 

n{n  —  l)(n  —  2) to  r  factors   n-rjf 

1x2x3 r 

Note.  In  finding  the  coefficient  of  the  (r  +  l)th  term,  write  down 
the  series  of  factors  1x2x3 r  for  the  denominator  of  the  coeffi- 
cient, then  write  over  this  series  the  factors  n(n  — l)(n  — 2),  etc.) 
writing  just  as  many  factors  in  the  numerator  as  there  are  in  the 
denominator. 

The  (r  +  l)th  term  in  the  expansion  of  (a  —  hy  is  the 
same  as  the  above  if  r  is  even,  and  the  negative  of  the 
above  if  r  is  odd. 


324  ALGEBKA.  ^^-^   <^ 


vT 


Find  the  eighth  term  of  {^~^^ •  ^  ^''^ 

Here  a  =  >,  6  =  ^,  n  =  10,  r  =  7. 


7    -,' 


V 


which  reduces  to  —  60  x^*. 

341.    A   trinomial   may   be   expanded  by  the   binomial 
theorem  as  follows : 

Expand(l4-2a;-a;'^)^ 

Put  2x~x^  =  z\ 

then  (l+z)3  =  l +  32  +  322 +  23_ 

Replace  z  with  2  a;  —  a^, 
.-.  (1  +  2x  -  a;2)3  =  1  +  3 (2a;  -  a;^)  +  3 (2a;  -  a;2)2  +  (2a;  -  ^2)3 
=  1  +  6a;  +  9x2  _  Ix^  -  9a;*  ^^x"  ~  ^. 

Exercise  114. 

1.  (l  +  2a7)^  3.    (2a;-3y)\  5.    A-^Y 

2.  (a; -3)1  4.    (2-a:)l  6.    (l-^^' 

7.  Find  the  fourth  term  of  (2:?;—  5yy^ 

8.  Find  the  seventh  term  of  (-  +  ^1  . 

9.  Find  the  twelfth  term  of  (a'  -  aa:)^^ 

10.  Find  the  eighth  term  of  (Sar^y  —  2rry^)^ 

11.  Findthemiddle  term  of /'~4-^Y. 

\y    V 

12.  Find  the  middle  term  of  (-  -  ^^ '° 


13.    Find  the  two  middle  terms  of 


BINOMIAL   THEOREM.  325 

14.  Find  the  rth  term  of  (2a  +  xf. 

15.  Find  the  rth  term  from  the  end  of  (2a  +  ^Y- 

16.  Find  the  (r  +  4)th  term  of  (a  +  xf. 

17.  Find  the  middle  term  of  (a  -{-  rp^". 

18.  Expand  (2a  +  xf^,  and  find  the  sum  of  the  terms  if 
•    a  =  1,  X  —  —  2. 

Expand : 

19.  ( Va  +  V^)^       24.  (Vm''-\-V^)\    29.  (Va  -  2 V^^ 

20.  (2a'^-|Vay.     25.  (2\/^-l/y.    30./"— -^/^Y- 

V  2^b)  V'C       2>)  \  2a) 

23.fJ^-4Y      28.f4^-3V6Y.33.f«4-^Y. 

342.  Binomial  Theorem,  Any  Exponent.  We  have  seen 
(§  338)  that  when  w  is  a  positive  integer  we  have  the 
identity 

(l  +  x)"^l  +  «x  +  ^^(^^+"^"7.^^^."3~^V  + 

We  proceed  to  the  case  of  fractional  and  negative  expo- 
nents. 


326  ALGEBKA. 

I.  Suppose  71  is  a  positive  fraction,  -?  We  may  assume 
that  ^' 

(l-i-xy  =  {A  +  Bx-j-  Cx'  +  Bx'  + )\  (1) 

provided  x  be  so  taken  that  the  series 

A  +  Bx+Cx'+Bx'i- 

is  convergent  (§  325). 

That  this  assumption  is  allowable  may  be  seen  as  follows : 

Expand  both  members  of  (1).     We  obtain 

^+P^+      1-2     "^^  1-2-3  ""^      ' 

and      A'  +  qA'-'Bx+^^^~^\A'-'jB'-{-qA'^-'C)x'  + 

1  *  Z 

In  the  first  k  coefficients  of  the  second  series  there  enter 

only  the  first  Jc  of  the  coefficients  A,  B,  C,  D, If,  then, 

we  equate  the  coefficients  of  corresponding  terms  in  the 
two  series  (§  330)  as  far  as  the  ^th  term,  we  shall  have  just 
h  equations  to  find  k  unknown  numbers  A,  B,  (7,  B,  ..... 
Hence  the  assumption  made  in  (1)  is  allowable. 

Comparing  the  two  first  terms  and  the  two  second  terms, 
we  obtain 

qA'~''B=p,  or  qB=p,     .'.  B=^- 
Extracting  the  ^-th  root  of  both  members  of  (1),  we  have 

(l  +  xy=li-?.x-j-Cx'-}~B:^-i- ,  (2) 

where  x  is  to  be  so  taken  that  the  series  on  the  right  is 
convergent. 


BINOMIAL    THEOREM.  327 

II.    Suppose  n  is  a  negative  number,  integral  or  frac- 
tional.    Let  n  =  —  m,  so  that  m  is  positive ;  then 

(i+.)«=(i+.)-.=^„. 

From  (2),  whether  m  is  integral  or  fractional,  we  may 
assume 

1 1 

(1  -^xY      1  +  mx  -{- ca^ -\- ds?  + ' 

By  actual  division  this  gives  an  equation  in  the  form 
(1  +  x)-"^  =  l-mx-\-Cx'-{-  Dx^  + (3) 

343.   It  appears  from  (2)  and  (3)  that  whether  n  be  inte- 
gral or  fractional,  positive  or  negative,  we  may  assume 

{l-\-xY  =  l-\-nx-{-  Cx"  +  Da?  + , 

provided  the  series  on  the  right  is  convergent. 
Squaring  both  members, 

(1  ■\-2x-\-  xy  =  1  +  271X  +  2Cx'  +  2Da?  + (1) 

-fTiV  -\-2nW. 
Also,  since 

we  have,  putting  2x-{-x'^  for  y, 

(1  4-  2a;  +  ^T  ^-l^n{2x-^x')^C{2x^  x'f 

+  D{2x^xy 

-=l-\-2nx  +  nx'  +4(7r'  + 

+  ^Cx'-\-SDx\  (2) 

Comparing  corresponding  coefficients  in  (1)  and  (2), 
w +  4(7=  2(7+ < 
4(7+8D  =  2Z>  +  2n(7. 

...  2(7=  w^- 71,  and  (7=''^^."^^ 

3i)  =  (n-2)(7,  and  D  =  ^ (^ " ,^X^ "  ^^ 
and  so  on. 


328  ALGEBRA. 

Hence,  whether  n  be  integral  or  fractional,  positive  or 
negative,  we  have 

/I   I     \n      II        I  ^(^—1)    2  r  n(n  —  l)(n—2)    ,  , 
•     (l  +  xy  =  l+nx-{ — \.2         "^         1-^3 ' 

if  X  is  so  taken  that  the  series  on  the  right  is  convergent. 

The  series  obtained  will  be  an  infinite  series  unless  n  is  a 
positive  integer  (§  336). 

344.  If  X  is  negative. 
Also,  a  x<.a, 

„/-,    ,      X  ,  n(n—  1) x^  ,        \ 
\  a  1'2      a^  J 

=  «"-[-  nal'-'x  +  ^^!^~^  a"-V  + ; 

if  a:>a, 

(a  +  xY  --=  (:r  +  ay  =  a;"  /^l  +  -Y' 

\          a;          1  •  2     a;^          / 
= :?;"  +  nax'''^  +  ^^^~-^^  aV"'  + 

345.  Examples. 

(1)  Expand  (l  +  x)^. 

i.  '  Z  1.  '  Z'  o 

=  l  +  ia;--^a;2  +  -A:^a;3_ 

'        3-6         3-6-9 

if  X  is  so  taken  that  the  series  is  convergent. 


BINOMIAL   THEOREM.  329 


(2)  Expand  (1  +  a;)"i 

^        3-6         3-6-9 


if  X  is  80  taken  that  the  series  is  convergent. 
(3)   Expand  - 


X 


vl  —  X 

if  X  is  so  taken  that  the  series  is  convergent. 

A  root  may  often  be  extracted  by  means  of  an  expansion. 

(4)  Extract  the  cube  root  of  344  to  six  decimal  places. 

.-.^344=      7(1.  ^)^ 

V        3V343J         1-2     V343y  J 

=  7(1  +  0.000971815  -  0.000000944) 
=  7.006796. 

(5)  Find  the  eishth  term  of  (x —  )    • 

.  V       WxJ 

Here  a  =  x,   b  = — -  =  — -,   n  =  —  -,   r  =  7. 


The  term  is 


Wx     4a.V  2 

l--f.--V---V 


i-a' 


1-2-3-4-5-6-7  X     4^a 

1-3-5-7-911-13-3T 


2  •  4  •  6  •  8  •  10  •  12  •  14  •  4'  •  x" 


9. 


Vi^d'-Saxf 


10.    'I L 


330  ALGEBRA. 

Exercise  115. 

Expand  to  four  terms : 

1.  (l  +  x)K  5.    (a'-x')\ 

2.  (l  +  x)^.  6.    (x'  +  XT/)-'^. 

3.  (a  +  x)K  7.    (2x-SyyK     ""*    \(l-3y)^ 

4.  (1-a;)-^  8.    \/r=:6^.  11.    (l4-^  +  ^^)§. 

12.  (l-a;  +  :c')l 

13.  Find  the  rth  term  of  (a  +  x)K 

14.  Find  the  rth  term  of  (a  —  x)~^. 

15.  Find  V65  to  five  decimal  places. 

16.  Find  VIjV  to  five  decimal  places. 

17.  Find  V 129  to  six  decimal  places. 

18.  Expand  (1  -  2a;  +  Zx^)~^  to  four  terms. 

19.  Find  the  coefficient  of  x^  in  the  expansion  of  (-*-  +  ^^/ 

(l  +  3a:y 

20.  By  means  of  the  expansion  of  (1  -f  x^  show  that  the 

limit  of  the  series 

14.1 1       ,       1X3  1x3x5  .     /K 

"^2     2x2^"^2x3x2^     2x3x4x2*"^ ' 


CHAPTER  XXVI. 
COMMON   LOGARITHMS. 

346.  If  the  natural  numbers  are  regarded  as  powers  of 
ten,  the  exponents  of  the  powers  are  the  Common  or  Briggs 
Logarithms  of  the  numbers.  If  A  and  £  denote  natural 
numbers,  a  and  b  their  logarithms,  then 

10«  =  ^,  10*  =  ^; 

or,  in  logarithmic  form, 

log  A  =  a,  log  B  =  b. 

347.  The  logarithm  of  a  product  is  found  by  adding  the 
logarithms  of  its  factors. 

For  AxB^  10«  X  10^  =  10«+^ 

Therefore,     log  (^  X  ^)  =  a  +  6  =  log  ^  +  log  B. 

348.  The  logarithm  of  a  quotient  is  found  by  subtracting 
the  logarithm  of*  the  divisor  from  that  of  the  dividend. 

For  A^^  =  lO<^-\ 

B     W 

Therefore,  \og  —  =  a  —  b  =  \ogA-  log  B. 

B 

349.  The  logarithm  of  a  power  is  found  by  multiplying 
the  logarithm  of  the  number  by  the  exponent  of  the  power. 

For  A''  =  (10«)"  =  10"". 

Therefore,  log  ^^  =  an  =  n  log  ^. 


332  ALGEBRA. 

350.  The  logarithm  of  the  root  of  a  number  is  found  by 
dividing  the  logarithm  of  the  number  by  the  index  of  the 
root. 

For  ^A  =  ^W=l(fr 

Therefore,  logV3  =  ^-i^^- 

n         n 

351.  The  logarithms  of  1,  10,  100,  etc.,  and  of  0.1,  0.01, 
0.001,  etc.,  are  integral  numbers.  The  logarithms  of  all 
other  numbers  are  fractions. 

Since      10"=      1,  10-1  (=J-^)      =0.1, 

10^=   10,  10-H=--TiTr)    =0.01, 

10^  =  100,  10-^  (=  y^Vn)  =  0.001, 

therefore   log      1  =  0,  log  0.1      =  —  1, 

log    10  =  1,  log  0.01    =-2, 

log  100  =  2,  log  0.001  =  -3. 

Also,  it  is  evident  that  the  common  logarithms  of  all 
numbers  between 

1  and       10  will  be      0  -f  a  fraction, 

10  and     100  will  be      1  -f  a  fraction, 

100  and  1000  will  be      2  +  a  fraction, 

1  and  0.1      will  be  —  1  +  a  fraction, 

0.1    and  0.01    will  be  —  2  +  a  fraction, 

0.01  and  0.001  will  be  -  3  +  a  fraction. 

352.  If  the  number  is  less  than  1,  the  logarithm  is  nega- 
tive (§  351),  but  is  written  in  such  a  form  that  the  fractional 
part  is  always  positive. 

353.  Every  logarfthm,  therefore,  consists  of  two  parts :  a 
positive  or  negative  integral  number,  which  is  called  the 
characteristic,  and  a  positive  proper  fraction,  which  is  called 


COMMON    LOGARITHMS.  333 

the  mantissa.  Thus,  in  the  logarithm  3.5218,  the  integral 
number  3  is  the  characteristic,  and  the  fraction  .5218  the 
mantissa.  In  the  logarithm  0.7825  —  2,  which  is  sometimes 
written  2.7825,  the  integral  number  —  2  is  the  character- 
istic, and  the  fraction  0.7825  is  the  mantissa. 

354.  If  the  logarithm  has  a  negative  characteristic,  it  is 
customary  to  change  its  form  by  adding  10,  or  a  multiple 
of  10,  to  the  characteristic,  and  then  indicating  the  sub- 
traction of  the  same  number  from  the  result.  Thus,  the 
logarithm  2.7825  is  changed  to  8.7825-10  by  adding  10 
to  the  characteristic  and  writing  —  10  after  the  result.  The 
logarithm  13.9273  is  changed  to  7.9273  -  20  by  adding  20 
to  the  characteristic  and  writing  —  20  after  the  result. 

355.  The  following  rules  are  derived  from  §  351 : 

Rule  1.  If  the  number  is  greater  than  1,  make  the 
characteristic  of  the  logarithm  one  unit  less  than  the  num- 
ber of  figures  on  the  left  of  the  decimal  point. 

Rule  2.  If  the  number  is  less  than  1,  make  the  charac- 
teristic of  the  logarithm  negative,  and  o?ie  unit  more  than 
the  number  of  zeros  between  the  decimal  point  and  the 
first  significant  figure  of  the  given  number. 

Rule  3.  If  the  characteristic  of  a  given  logarithm  is 
positive,  make  the  number  of  figures  in  the  integral  part  of 
the  corresponding  number  one  more  than  the  number  of 
units  in  the  characteristic. 

Rule  4.  If  the  characteristic  is  negative,  make  the  num- 
ber of  zeros  between  the  decimal  point  and  the  first  signifi- 
cant figure  of  the  corresponding  number  one  less  than  the 
number  of  units  in  the  characteristic. 

Thus,  the  characteristic  of  log  7849.27  is  3 ;  the  character- 
istic of  log  0.037  is  -  2  =  8.0000  -  10.   If  the  characteristic 


334  ALGEBRA. 

is  4,  the  corresponding  number  has  five  figures  in  its  integral 
part.  If  the  characteristic  is  —  3,  that  is,  7.0000  —  10,  the 
corresponding  fraction  has  two  zeros  between  the  decimal 
point  and  the  first  significant  figure. 

356.  The  mantissa  of  the  common  logarithm  of  any  inte- 
gral number,  or  decimal  fraction,  depends  only  upon  the 
digits  of  the  number,  and  is  unchanged  so  long  as  the 
sequence  of  the  digits  remains  the  same. 

For  changing  the  position  of  the  decimal  point  in  a 
number  is  equivalent  to  multiplying  or  dividing  the  num- 
ber by  a  power  of  10.  Its  common  logarithm,  therefore, 
will  be  increased  or  diminished  by  the  exponent  of  that 
power  of  10 ;  and  since  this  exponent  is  integral,  the  man- 
tissa, or  decimal  part  of  the  logarithm,  will  be  unaffected. 

Thus,    27196  =  10*-"^^  2.7196  =  10"*=^^ 

2719.6  =  10'•*'*^        0.27196  =  lO^-*^^^^  -^'', 
27.196  =  10'-'''',    0.0027196  =  lo^-^^^^-^". 

One  advantage  of  using  the  number  ten  as  the  base  of  a 
system  of  logarithms  consists  in  the  fact  that  the  mantissa 
depends  only  on  the  sequence  of  digits,  and  the  characteristic 
on  the  position  of  the  decimal  point. 

357.  In  simplifying  the  logarithm  of  a  root  the  equal 
positive  and  negative  numbers  to  be  added  to  the  logarithm 
should  be  such  that  the  resulting  negative  number,  when 
divided  by  the  index  of  the  root,  gives  a  quotient  of  —  10. 

Thus,  if  the  log  0.002^  =  J  of  (7.3010  -  10),  the  expres- 
sion J  of  (7.3010  —  10)  may  be  put  in  the  form  I  of 
(27.3010  -  30),  which  is  9.1003  -  10,  since  the  addition 
of  20  to  the  7,  and  of—  20  to  the  — 10,  produces  no  change 
in  the  vahie  of  the  logarithm. 


COIiMON    LOGARITHMS.  335 

Exercise  116. 

Given:  log 2  =0.3010;  log 3  =  0.4771;  log 5  =  0.6990; 
log  7  =  0.8451. 

Find  the  common  logarithms  of  the  following  numbers 
by  resolving  the  numbers  into  factors,  and  taking  the  sum 
of  the  logarithms  of  the  factors. 

1.  log  35.     5.    log  12.         9.    log 0.05.       13.    log  1.75. 

2.  log  9.       6.    log  60.       10.    log  12.5.       14.    log  105. 

3.  log  8.       7.    log  75.       11.    log  1.25.       15.    log  0.0105. 

4.  log  49.     8.    log  7.5.      12.    log  37.5.       16.    log  1.05. 
Find  the  common  logarithms  of  the  following  : 


17. 

7^ 

20. 

5i 

23. 

2l 

26. 

3xr. 

29. 

5l 

18. 

31 

21. 

3*. 

24. 

5*. 

27. 

7l 

30. 

7-V. 

19. 

71 

22. 

7^. 

25. 

3l 

28. 

3^. 

31. 

21^. 

358.    The   logarithm  of  the  reciprocal  of  a  number  is 
called  the  cologarithm  of  the  number. 
If  A  denote  any  number,  then 

colog  ^  =  log  i  =  log  1  -  log  ^  (§  348)  =  -  log  ^  (§  351). 

Hence,  the  cologarithm  of  a  number  is  equal  to  the  log- 
arithm of  the  number  with  the  minus  sign  prefixed,  which 
sign  affects  the  entire  logarithm,  both  characteristic  and 
mantissa. 

In  order  to  avoid  a  negative  mantissa  in  the  cologarithm, 
it  is  customary  to  substitute  for  —  log^  its  equivalent 
(10  -  log  A)  -  10. 

Hence,  the  cologarithm  of  a  number  is  found  by  subtract- 
ing the  logarithm  of  the  number  from  10,  and  then  annexing 
—  10  to  the  remainder. 


336  ALGEBRA. 

The  best  way  to  perforin  the  subtraction  is  to  begin  on 
the  left  and  subtract  each  figure  of  log^  from  9  until  we 
reach  the  last  significant  figure,  which  must  be  subtracted 
from  10. 

If  log  A  is  greater  in  absolute  value  than  10  and  less 
than  20,  then  in  order  to  avoid  a  negative  mantissa,  it  is 
necessary  to  write  —  log  A  in  the  form  (20  —  log  A)  —  20. 
So  that,  in  this  case,  colog  A  is  found  by  subtracting  log  A 
from  20,  and  then  annexing  —  20  to  the  remainder. 

(1)  Find  the  cologarithm  of  4007. 

&  10-10 

Given:  log  4007=    3.6028 

Therefore  colog  4007=    6.3972-10 

(2)  Find  the  cologarithm  of  103992000000. 

20  -20 

Given  : ,  log  103992000000  =  11.0170 

Therefore,         colog  103992000000  =    8.9830-20 

If  the  characteristic  of  log  A  is  negative,  then  the  subtra- 
hend, —  10  or  —  20,  will  vanish  in  finding  the  value  of 
colog  A. 


(3)   Find  the 

col( 

Dgarithm  of  0.004007. 

10 

-10 

Given : 

log  0.004007  = 
colog  0.004007  = 

7.6028- 

-10 

Therefore, 

2.3972 

By  using  cologarithms  the  inconvenience  of  subtracting 
the  logarithm  of  a  divisor  is  avoided.  For  dividing  by  a 
number  is  equivalent  to  multiplying  by  its  reciprocal. 
Hence,  instead  of  subtracting  the  logarithm  of  a  divisor,  its 
cologarithm  may  be  added. 


(1)   Find  the  logarithm  of 


COMMON    LOGARITHMS.  337 

5 


0.002 


log  --A_  =  log  5  +  coloff  0.002 
^0.002         ^  ^ 

log  5  =  0.6990 

colog  0.002  =  2.0990 

log  quotient  =  3.3980 

(2)   Find  the  logarithm  of  ^- 

A 

Iog^  =  log0.07  +  colog23. 

log  0.07  =  8.8451 -10 
colog  2=^  =  (10  -  3  log  2)  -  10  =  9.0970  -  10 


log  quotient  ==  7.9421  -  10 


"Exercise  117. 


Given:  log 2  =  0.3010;  log 3  =  0.4771;  log 5  =  0.6990 
log  7  =  0.8451 ;  log  11  =  1.0414. 

Find  the  logarithms  of  the  following  quotients : 


3.  2 

5 

4.  ^ 

7 

5.  5 
6. 


2 
5' 

7. 

5 
3" 

13. 

0.05 
3 

19. 

0.05 
0.003 

25. 

0.022 
3» 

2 

i 

8. 

5 
2* 

14. 

0.005 

2 

20. 

0.007 
0.02 

26. 

33 

0.022 

3 
5' 

9. 

7 
3' 

15. 

0.07 
5 

21. 

0.02 
0.007 

27. 

73 
0.022 

3 

i 

10. 

7 

i 

16. 

5 
0.07 

22. 

0.005 
0.07 

28. 

0.07' 
0.003» 

5 

7" 

11. 

3 

i 

17. 

3 
0.007 

23. 

0.03 

7 

29. 

0.0052 
73 

7 
5 

12. 

7 
0.5' 

18. 

0.003 

7 

24. 

0.0007 
0.2 

30. 

,73 
0.0052 

^^^' 


1 


338  ALGEBRA. 

359.  Tables.  A  table  oi  four-place  common  logarithms 
is  given  on  pages  340  and  341,  which  contains  the  common 
logarithms  of  all  numbers  under  1000,  the  decimal  point 
and  characteristic  being  omitted.  The  logarithms  of  single 
digits,  1,  8,  etc.,  will  be  found  at  10,  80,  etc. 

Tables  containing  logarithms  of  more  places  can  be  pro- 
cured, but  this  table  will  serve  for  many  practical  uses,  and 
will  enable  the  student  to  use  tables  of  five-place,  seven- 
place,  and  ten-place  logarithms,  in  work  that  requires 
greater  accuracy. 

In  working  with  a  four-place  table,  the  numbers  corre- 
sponding to  the  logarithms,  that  is,  the  antilogarithms,  as 
they  are  called,  may  be  carried  to  four  significant  digits. 

360.  To  find  the  Logarithm  of  a  Number  in  this  Table. 

(1)  Suppose  it  required  to  find  the  logarithm  of  65.7. 
In  the  column  headed  "  N"  look  for  the  first  two  significant 
figures,  and  at  the  top  of  the  table  for  the  third  significant 
figure.  In  the  line  with  65,  and  in  the  column  headed  7, 
is  seen  8176.  To  this  number  prefix  the  characteristic  and 
insert  the  decimal  point.     Thus, 

log65.7  =  1.8176. 

(2)  Suppose  it  is  required  to  find  the  logarithm  of  20347. 
In  the  line  with  20,  and  in  the  column  headed  3,  is  seen 
3075;  also  in  the  line  with  20,  and  in  the  4  column,  is  seen 
3096,  and  the  difi^erence  between  these  two  is  21.  The  dif- 
ference between  20300  and  20400  is  100,  and  the  difference 
between  20300  and  20347  is  47.  Hence,  yVy  of  21  =  10, 
nearly,  must  be  added  to  3075 ;  that  is, 

log  20347 -4.3085. 

(3)  Suppose  it  is  required  to  find  the  logarithm  of 
0.0005076.  In  the  line  with  50,  and  in  the  7  column,  is 
seen  7050;  in  the  8  column,  7059:  the  difference  is  9.    The 


COMMON    LOGARITHMS.  339 

difference  between  5070  and  5080  is  10,  and  the  difference 
between  5070  and  5076  is  6.  Hence,  y^^  of  .9  =  5  must  be 
added  to  7050 ;  that  is, 

log 0.0005076  =  6.7055  -  10. 
361.   To  find  a  Number  when  its  Logarithm  is  given. 

(1)  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  1.9736. 

Look  for  9736  in  the  table.  In  the  column  headed  "N," 
and  in  the  line  with  9736,  is  seen  94,  and  at  the  head  of 
the  column  in  which  9736  stands  is  seen  1.  Therefore, 
write  941,  and  insert  the  decimal  point  as  the  characteristic 
directs;  that  is,  the  number  required  is  94.1. 

(2)  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  3.7936. 

Look  for  7936  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
7931  and  7938 ;  their  difference  is  7,  and  the  difference  be- 
tween 7931  and  7930  is  5.  Therefore,  ^  of  the  difference 
between  the  numbers  corresponding  to  the  mantissas,  7931 
and  7938,  must  be  added  to  the  number  corresponding  to 
the  mantissa  7931. 

The  number  corresponding  to  the  mantissa  7938  is  6220. 

The  number  corresponding  to  the  mantissa  7931  is  6210. 

The  difference  between  these  numbers  is  10, 
and  6210  +  ^  of  10  -  6217. 

Therefore,  the  number  required  is  6217. 

(3)  Suppose  it  is  required  to  find  the  number  of  which 
the  logarithm  is  7.3882  -  10. 

Look  for  3882  in  the  table.  It  cannot  be  found,  but  the 
two  adjacent  mantissas  between  which  it  lies  are  seen  to  be 
3874  and  3892 ;  the  difference  between  the  two  mantissas 
is  18,  and  the  difference  between  3874  and  the  given  man- 
tissa 3882  is  8. 


340 


ALGEBRA. 


\ 


N 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

lO 

11 
12 
13 
14 

0000 
0414 
0792 
1139 
1461 

0043 
0453 
0828 
1173 
1492 

0086 
0492 
0864 
1206 
1523 

0128 
0531 
0899 
1239 
1553 

0170 
0569 
0934 
1271 
1584 

0212 
0607 
0969 
1303 
1614 

0253 
0645 
1004 
1335 
1644 

0294 
0682 
1038 
1367 
1673 

0334 
0719 
1071 
1399 
1703 

0374 
0755 
1106 
1430 
1732 

15 

16 
17 
18 
1.9 

1761 
2041 
2304 
2553 

2788 

1790 
2068 
2330 
2577 
2810 

1818 
2095 
2355 
2601 
2833 

1847 
2122 
2380 
2625 
2856 

1875 
2148 
2405 
2648 
2878 

1903 
2175 
2430 
2672 
2900 

1931 
2201 
2455 
2695 
2923 

1959 
2227 
2480 
2718 
2945 

1987 
2253 
2504 

2742 
2967 

2014 
2279 
2529 
2765 
2989 

20 

21 
22 
23 
24 

3010 

^?,?,?, 
3424 
3617 
3802 

3032 
3243 
3444 
3636 
3820 

3054 
3263 
3464 
3655 

3838 

3075 
3284 
3483 
3674 
3856 

3096 
3304 
3502 
3692 

3874 

3118 
3324 
3522 
3711 
3892 

3139 
3345 
3541 
3729 
3909 

3160 
3365 
3560 
3647 
3927 

3181 
3385 
3579 
3766 
3945 

3201 
3404 
3598 

3784 
3962 

25 

26 
27 
28 
29 

3979 
4150 
4314 
4472 
4624 

3997 
4166 
4330 
4487 
4639 

4014 
4183 
4346 
4502 
4654 

4031 
4200 
4362 
4518 
4669 

4048 
4216 
4378 
4533 
4683 

4065 
4232 
4393 
4548 
4698 

4082 
4249 
4409 
4564 
4713 

4099 
4265 
4425 
4579 

4728 

4116 
4281 
4440 
4594 
4742 

4133 

4298 
4456 
4609 

4757 

30 

31 
32 
33 
34 

4771 
4914 
5051 
5185 
5315 

4786 
4928 
5065 
5198 
5328 

4800 
4942 
5079 
5211 
5340 

4814 
4955 
5092 
5224 
5353 

4829 
4969 
5105 
5237 
5366 

4843 
4983 
5119 
5250 
5378 

4857 
4997 
5132 
5263 
5391 

4871 
5011 
5145 
5276 
5403 

4886 
5024 
5159 
5289 
5416 

4900 
5038 
5172 
5302 
5428 

35 

36 
37 
38 
39 

5441 
5563 
5682 
5798 
5911 

5453 
5575 
5694 
5809 
5922 

5465 
5587 
5705 
5821 
5933 

5478 
5599 
5717 
5832 
5944 

5490 
5611 
5729 
5843 
5955 

5602 
5623 
5740 
5855 
5966 

5514 
5635 
5752 
5866 
5977 

5527 
5647 
5763 

5877 
5988 

5539 
5658 
5775 
5888 
5999 

5551 
5670 
5786 
5899 
6010 

40 

41 
42 
43 
44 

6021 
6128 
6232 
6335 
6435 

6031 
6138 
6243 
6345 
6444 

6042 
6149 
6253 
6355 
6454 

6053 
6160 
6263 
6365 
6464 

6064 
6170 
6274 
6375 
6474 

6075 
6180 
6284 
6385 
6484 

6085 
6191 
6294 
6395 
6493 

6096 
6201 
6304 
6405 
6503 

6107 
6212 
6314 
6415 
6513 

6117 
6222 
6325 
6425 
6522 

45 

46 
47 
48 
49 

6532 
6628 
6721 
6812 
6902 

6542 
6637 
6730 
6821 
6911 

6551 
6646 
6739 
6830 
6920 

6561 
6656 
6749 
6839 
6928 

6571 
6665 
6758 
6848 
6937 

6580 
6675 
6767 
6857 
6946 

6590 
6684 
6776 
6866 
6955 

6599 
6693 
6785 
6875 
6964 

6609 
6702 
6794 

6884 
6972 

6618 
6712 
6803 
6893 
6981 

50 

51 
52 
53 
54 

6990 
7076 
7160 
7243 
7324 

6998 
7084 
7168 
7251 
7332 

7007 
7093 
7177 
7259 
7340 

7016 
7101 
7185 
7267 
7348 

7024 
7110 
7193 

7275 
7356 

7033 
7118 
7202 

7284 
7364 

7042 
7126 
7210 
7292 

7372 

7050 
7135 
7218 
7300 

7380 

7059 
7143 
7226 
7308 
7388 

7067 
7152 
7235 
7316 
7396 

COMMON   LOGARITHMS. 


341 


N 

55 

56 
57 
58 
59 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

7404 
7482 
7559 
7634 
7709 

7412 
7490 
7566 
7642 
7716 

7419 
7497 
7574 
7649 
7723 

7427 
7505 
7582 
7657 
7731 

7435 
7513 
7589 
7664 
7738 

7443 
7520 
7597 
7672 
7745 

7451 
7528 
7604 
7679 

7752 

7459 
7536 
7612 
7686 
7760 

7466 
7543 
7619 
7694 
7767 

7474 
7551 
7627 
7701 

7774 

60 

61 
62 
63 
64 

7782 
7853 
7924 
7993 
8062 

7789 
7860 
7931 
8000 
8069 

7796 
7868 
7938 
8007 
8075 

7803 
7875 
7945 
8014 
8082 

7810 
7882 
7952 
8021 
8089 

7818 
7889 
7959 
8028 
8096 

7825 
7896 
7966 
8035 
8102 

7832 
7903 
7973 
8041 
8109 

7839 
7910 
7980 
8048 
8116 

7846 
7917 
7987 
8055 
8122 

65 

66 
67 
68 
69 

8129 
8195 
8261 
8325 
8388 

8136 
8202 
8267 
8331 
8395 

8142 
8290 
8274 
8338 
8401 

8149 
8215 
8280 
8344 
8407 

8156 
8222 
8287 
8351 
8414 

8162 
8228 
8293 
8357 
8420 

8169 
8235 
8299 
8363 
8426 

8176 
8241 
8306 
8370 
8432 

8182 
8248 
8312 
8376 
8439 

8189 
8254 
8319 
8382 
8445 

70 

71 
72 
73 
74 

8451 
8513 
8573 
8633 
8692 

8457 
8519 
8579 
8639 
8698 

8463 
8525 
8585 
8645 
8704 

8470 
8531 
8591 
8651 
8710 

8476 
8537 
8597 
8657 
8716 

8774 
8831 
8887 
8943 
8998 

8482 
8543 
8603 
8663 
8722 

8488 
8549 
8609 
8669 
8727 

8494 
8555 
8615 
8675 
8733 

8500 
8561 
8621 
8681 
8739 

8506 
8567 
8627 
8686 
8745 

75 

76 
77 
78 
79 

8751 
8808 
8865 
8921 
8976 

8756 
8814 
8871 
8927 
8982 

8762 
8820 
8876 
8932 
8987 

8768 
8825 
8882 
8938 
8993 

8779 
8837 
8893 
8949 
9004 

8785 
8842 
8899 
8954 
9009 

8791 
8848 
8904 
8960 
9015 

8797 
8854 
8910 
8965 
9020 

8802 
8859 
8915 
8971 

9025 

80 

81 
82 
83 
84 

9031 
9085 
9138 
9191 
9243 

9036 
9090 
9143 
9196 
9248 

9042 
9096 
9149 
9201 
9253 

9047 
9101 
9154 
9206 
9258 

9053 
9106 
9159 
9212 
9263 

9058 
9112 
9165 
9217 
9269 

9063 
9117 
9170 
9222 
9274 

9069 
9122 
9175 
9227 
9279 

9074 
9128 
9180 
9232 
9284 

9079 
9133 
9186 
9238 
9289 

85 

86 
87 
88 
89 

9294 
9345 
9395 
9445 
9494 

9299 
9350 
9400 
9450 
9499 

9304 
9355 
9405 
9455 
9504 

9309 
9360 
9410 
9460 
9509 

9315 
9365 
9415 
9465 
9513 

9320 
9370 
9420 
9469 
9518 

9325 
9375 
9425 
9474 
9523 

9330 
9380 
9430 
9479 
9528 

9335 
9385 
9435 
9484 
9533 

9340 
9390 
9440 
9489 
9538 

90 

91 
92 
93 
94 

9542 
9590 
9638 
9685 
9731 

9547 
9595 
9643 
9689 
9736 

9552 
9600 
9647 
9694 
9741 

9557 
9605 
9652 
9699 
9745 

9562 
9609 
9657 
9703 
9750 

9566 
9614 
9661 
9708 
9754 

9571 
9619 
9666 
9713 
9759 

9576 
9624 
9671 
9717 
9763 

9581 
9628 
9675 
9722 
9768 

9586 
9633 
9680 
9727 
9773 

95 

96 
97 
98 
99 

9777 
9823 
9868 
9912 
9956 

9782 
9827 
9872 
9917 
9961 

9786 
9832 
9877 
9921 
9965 

9791 
9836 
9881 
9926 
9969 

9795 
9841 
9886 
9930 
9974 

9800 
9845 
9890 
9934 
9978 

9805 
9850 
9894 
9939 
9983 

9809 
9854 
9899 
9943 
9987 

9814 
9859 
9903 
9948 
9991 

9818 
9863 
9908 
9952 
9996 

342  ALGEBRA. 

The  number  corresponding  to  the  mantissa  3892  is  2450. 
The  number  corresponding  to  the  mantissa  3874  is  2440. 
The  difference  between  these  numbers  is  10, 
and  2440  +  r\  of  10  =  2444. 

Therefore,  the  number  required  is  0.002444. 

Exercise  118. 
Find  from  the  table  the  logarithms  of: 


1. 

999. 

4.  90801. 

7.  0.00987. 

10.  7.0699, 

2. 

9901. 

5.  10001. 

8.  0.87701. 

11.  0.0897. 

3. 

5406. 

6.  10010. 

9.  1.0001. 

12.  99.778. 

Find  antilogarithms  to  the  following  common  logarithms  : 

13.  2.5310.  15.    9.8800-10.  17.    7.0216-10. 

14.  1.9484.  16.    0.2787.  18.    8.6580-10. 

362.   Examples. 

(1)  Find  the  product  of  908.4  X  0.05392  x  2.117. 

log       908.4  =  2.9583 
log    0.05392  =  8.7318-10 
log       2.117  =  0.3257 

2.0158  =  log  103.7. 

When  any  of  the  factors  are  negative,  find  their  logarithms  with- 
out regard  to  the  signs ;  write  —  after  the  logarithm  that  corresponds 
to  a  negative  number.  If  the  number  of  logarithms  so  marked  is 
odd,  the  product  is  negative ;  if  even,  the  product  is  positive. 

/o\  T?-  A  ^i.         +-4.    P  -  8.3709  X  834.637 

(2)  Find  the  quotient  of : 

^  ^  ^  7308.946 

log     8.3709  =  0.9227 
log   834.637  =  2.9215  + 

colog  7308.946  =  6.1362  -  10  + 

9.9804 -10  =  log -0.9558. 


COMMON    LOGARITHMS.  343 

(3)  Find  the  cube  of  0.0497. 

log  0.0497  =  8.6964 -10 
Multiply  by  3,  3 

6.0892 -10  =  log  0.0001228. 

(4)  Find  the  fourth  root  of  0.0086^ 

log  0.00862=    7.9355-10 
Add  30 -30,  30-30 

Divide  by  4,  4)37.9355-40 

9.4839 -10  =  log  0.3047. 

(5)  Fmd  the  value  of  ^3.1416x4771.21x27183^. 

30.103^X0.4343^X69.897* 

log   3.1416  =  0.4971  =0.4971 

log  4771.21  =  3.6786  =3.6786 

ilog   2.7183  =  0.4343-^2        =0.2172 
4  colog   30.103  =  4  (8.5214  -  10)  =  4.0856  -  10 
^colog   0.4343  =  0.3622-4-2        =0.1811 
4  colog   69.897  =  4(8.1555  -  10)  =  2.6220  -  10 

11.2816  -  20 
30  -  30 


5)41.2816-50 
8.2563  -  10 

=  log  0.01804. 

363.  An  exponential  equation,  that  is,  an  equation  in  which 
the  exponent  involves  the  unknown  number,  is  easily  solved 
by  Logarithms. 

Ex.  Find  the  value  of  x  in  8P  =  10. 

81*  =  10. 
.-.  log  (81==)  =  log  10, 
a;  log  81  =  log  10, 

loglO^L0000^0  524 
log  81      1.9085 


344 


ALGEBRA. 


Exercise  119. 
Find  by  logarithms  the  following  : 

1.  948.76x0.043875.        5.    7564  x  (- 0.003764). 

2.  3.4097x0.0087634.      6.   3.7648  x  (- 0.083497). 

3.  830.75x0.0003769.       7.   -  5.840359  X  (-0.00178). 

8.   -  8945.07  X  73.846. 

0.07654 


4.   8.4395x0.98274 
70654 


9. 
10. 
11. 
12. 

13.    - 

19.  6.05^ 

20.  1.05r. 

21.  1.17681 


54013 

58706 
93078' 

8.32165 
0.07891' 

650C9 
90761' 

7.652 
-  0.06875' 

26.  (i{)\ 

27.  (10|)\ 

28.  my. 


14. 


15. 


16. 


17. 


83.947  X  0.8395 

7564  X  0.07643 
8093  X  0.09817' 

89  X  753  X  0.0097 
36709  X  0.08497  ' 

413  X  8.17  X  3182 
915x728x2.315' 


18     212  X(- 6.12)  X(- 2008) 
365  X  (- 531)  X  2.576 

33.  (8f)^•^        40.  8.1904^. 

34.  (5f|)°-"^    41.  0.17643^ 


22.  1.3178^^     29.  (f|i)«. 

23.  0.78765^    30.  (7j\y- 

24.  0.69P. 

25.  (i^y\ 

47.  5^  =  20. 


35.  7i  42.  2.5637rT 

36.  lli  43.   (m)K 

37.  783^.  44.  (^AVe)'. 

31.  (3|i)*-^l     38.  8379^'^.       45.  (9fi)l 

32.  (It^^-)^'-^      39.  906.80^     46.  (llfl)^. 
48.  (1.3)^^  =  2.1.  49.  (0.9)^  =  |. 


COMMON    LOGARITHMS.  345 


50. 


51. 


52. 


53. 


54. 


55. 


56. 


57. 


58. 


510.0075433^  X  78.343  X  8172.43  x  0.00052 
\      64285^  X  154.27*  X  0.001  X  586.79* 

5/         15.832--^  X  5793.6^  X  0.78426 

^^0.000327^  X  768.94^  x  3015.3  X  0.007*' 

5/7.1895  X  4764.2^  X  0.00326^ 
\0.00048953  X  457'  X  5764.4^' 

3.1416  X  4771.21  x  2.7183* 
30.103*  X  0.4343*  X  69.897*' 


4 


J0.0327P  X  53.429  X  0.77542^ 
\         32.769  X  0.000371* 

3I 732.056'  X  0.0003572*  x  89793 
\42.2798-'^  X  3.4574  x  0.0026518^' 

3|7932  X  0.00657  X  0.80464 
\        0.03274  X  0.6428 

7.1206  X  VO.  13274  X  0.057389 


4 
{ 


VO.43468  X  17.385  X  VO.0096372 

3.075526'  X  5771.2*  x  0.0036984^  X  7.74  )i 
72258  X  327.93'  x  86.97^  i 


Note.  It  is  assumed  in  this  chapter  that  the  index  laws  which 
have  been  established  for  commensurable  exponents  hold  good  for 
incommensurable  exponents.  For  the  proof  see  Wentworth's  College 
Algebra,  g  26i,  page  216. 

Any  positive  number,  except  1,  may  be  selected  as  the  base ;  and 
to  the  base  selected  there  corresponds  a  system  of  logarithms. 


CHAPTER  XXVII. 

INTEREST  AND   ANNUITIES. 

364.  Simple  Interest. 

If  the  principal  is  represented  by  P, 

the  interest  on  $  1  for  one  year  by  r, 

the  amount  of  $1  for  one  year  by  i?, 

the  number  of  years  by  n, 

the  amount  of  P  for  n  years  by  A, 

Then  i?  =  1  +  r. 

Simple  interest  on  P  for  a  year      =  Pr, 
Amount  of  P  for  a  year  =  PR, 

Simple  interest  on  P  for  n  years    =  Pnr, 
Amount  of  P  for  n  years  =  P(l  +  nr), 

that  is,  A  =  P(l  +  nr), 

365.  When  any  three  of  the  quantities  A,  P,  ?2,  r  are 
given,  the  fourth  may  be  found. 

Required  the  rate  when  $  500  in  4  years  at  simple  interest 
amounts  to  $610. 

r  is  required,  A,  P,  n  are  given. 
A  =  P{1+  nr), 
or  A-=  P+  Pnr. 

.-.  Pnr  =  A-P. 

A-P     610  -  500 


Pn  2000 


0.055. 


INTEREST    AND   ANNUITIES.  347 

366.  Since  P  will  in  n  years  amount  to  ^,  it  is  evident 
that  P  at  the  present  time  may  be  considered  equivalent  in 
value  to  A  due  at  the  end  of  n  years ;  so  that  P  may  be 
regarded  as  the  present  worth  of  a  given  future  sum  A. 

Find  the  present  worth  of  $600,  due  in  2  years,  the  rate 
of  interest  being  6  per  cent. 

A  =  P(l  +V). 
.    D        A     _    1600    _  1535  7] 


1  +  nr     1  +  0.12 

367.   Compound  Interest. 

I.  When  compound  interest  is   reckoned  payable  an- 
nually, 

The  amount  of  P  dollars  in 

1  year  is       P(l  +  r)  or  PP, 

2  years  is  PP  (1  +  r)  or  PR\ 
n  years  is  PR*^. 

That  is,  A  =  PR'. 

Hence,  also,  -^^~B^' 

II.  When  compound  interest  is  payable  semi-annually, 
The  amount  of  P  dollars  in 

^year  isP^l-f|\ 
1  year  is  P[  1  4-^) » 
M years  is  P[  1  +M  • 

That  is,  ^^p/'l+iy" 

III.  When  the  interest  is  payable  quarterly^ 


348  ALGEBRA. 

IV.  When  the  interest  is  payable  monthly, 

V.  When  interest  is  payable  q  times  a  year, 

Find  the  present  worth  of  $500,  due  in  4  years,  at  5  per 
cent  compound  interest. 

^  =  P(l  +  rf. 

:.  P=      ^      =  1^  =  $411.36. 
(1  +  rf     (1.05)*     * 

368.  Sinking  Funds,  If  the  sum  set  apart  at  the  end  of 
each  year  to  be  put  at  compound  interest  is  represented 
by  8,  then 

The  sum  at  the  end  of  the 
first  year      =  S, 
second  year  =  8-\-  SB, 
third  year    =S-{-SE-i-SB\ 

nth  year       =Si-SB-\-  SB'  + +  SB"'-'. 

That  is,  the  amount  A  =  S-i- SB  + SB^  + +  SB""-'. 

.-.  AB  =  SB  +  SB'  +  SB'  + +  SB\ 

.-.  AB-A=-SB''~S. 

.'.A==^(^^-^\ 
B~l 

A=^(^-^\ 
r 

(1)  If  $10,000  be  set  apart  annually,  and  put  at  6  per 
cent  compound  interest  for  10  years,  what  will  be  the 
amount  ? 

_4  -_  ^JR""  -  1)  ^  $10,000(1.0610  -  1) 
r  0.06 

By  logarithms  the  amount  is  found  to  be  1131,740  (nearly). 


INTEREST   AND   ANNUITIES.  349 

(2)  A  county  owes  $60,000.  What  sum  must  be  set 
apart  annually,  as  a  sinking  fund,  to  cancel  the  debt  in  10 
years,  provided  money  is  worth  6  per  cent  ? 

c»        Ar        $60,000x0.06     a>.p:Kc,        ,, 

Note.  The  amount  of  tax  required  yearly  is  $  3600  for  the  interest 
and  $4555  for  the  sinking  fund;  that  is,  $8155. 

369.  Annuities.  A  sum  of  money  that  is  payable  yearly, 
or  in  parts  at  fixed  periods  in  the  year,  is  called  an  annuity. 

To  find  the  amount  of.  an  unpaid  annuity  when  the  inter- 
est, time,  and  rate  per  cent  are  given. 
The  sum  due  at  the  end  of  the 
first  year  .    =  S, 
second  year  ~  8-\-  8R, 
third  year    =-8+8R^8R, 
wthyear       =  8+ 8R-\- 8^:" -]- -\- 8R''\ 

That  is,  A=^^^-^\ 

r 

An  annuity  of  $  1200  was  unpaid  for  6  years.    What  was 

the  amount  due  if  interest  is  reckoned  at  6  per  cent  ? 

jl  ■_  8{Rn  - 1)  _  $  1200(1.06g  - 1) . .  ^  g3^Q 

r  0.06  ^ 

370.  To  find  the  present  worth  of  an  annuity  vjhen  the 
time  it  is  to  continue  and  the  rate  per  cent  are  given. 

Let  P  denote  the  present  worth.  Then  the  amount  of 
P  for  71  years  will  be  equal  to  A,  the  amount  of  the  annuity 
for  n  years. 

Therefore  for  n  years 

A  =  P{l-{-rY  =  PBr,  §367 

and  A  =^  ^^^  ~  ^X  §369 

P~l 


350  ALGEBRA. 


This  equation  may  be  written 


B-l        R'        R 
As  n  increases,  the  expression 


^.C-i) 


(-i) 


approaches  1.     Therefore,  if  the  annuity  \^  perpetual^ 

p^     8     _8 
R-1      r 

(1)  Find  the  present  worth  of  an  annual  pension  of 
$  105  for  5  years,  at  4  per  cent  interest. 

1.045      1.04-1      ^        ^         ^^ 

(2)  Find  the  present  worth  of  a  perpetual  scholarship 
that  pays  $300  annually,  at  6  per  cent  interest. 

p  =  ^=l^  =  $5000. 
r      0.06 

371.  To  find  the  present  worth  of  an  annuity  that  begins 
in  a  given  number  of  years,  when  the  time  it  is  to  continue 
and  the  rate  per  cent  are  given. 

Let  p  denote  the  number  of  years  before  the  annuity 
begins,  and  q  the  number  of  years  the  annuity  is  to  con- 
tinue. 

Then  the  present  worth  of  the  annuity  to  the  time  it 
terminates  is 

8       B'+'  - 1 
i2^+*       R-l' 


INTEREST    AND   ANNUITIES.  351 

and  the  present  worth  of  the  annuity  to  the  time  it  begins  is 

R^      R-\ 
Hence, 

/  s       ii!-+»-l\     (8      B?-\\ 

■    p-    S    yR'-l 
i?>+'      R-l 

If  the  annuity  is  to  begin  at  the  end  of  p  years,  and  to 
be  perpetual,  the  formula 

J?^+«  ^  R-l 

becomes  R=^,z:tt. — tt^  — ;:^^ 

R'{R-\)         R" 

And  since  — - — ■  approaches  1  (§  370), 
R^ 

p  ^ 


R\R--\) 

(1)  Find  the  present  worth  of  an  annuity  of  $5000,  to 
begin  in  6  years,  and  to  continue  12  years,  at  6  per  cent 
interest. 

^15000      roeii-i^^  29,550. 
1.06^8  0.06 

(2)  Find  the  present  worth  of  a  perpetual  annuity  of 
$  1000,  to  begin  in  3  years,  at  4  per  cent  interest. 

D  ^        -      $10^0      _  122,225. 


Rp^R-l)     1.013x0.04 


352  ALGEBRA. 

372.    To  find  the  annuity  when  the  present   worth,  the 
time,  and  the  rate  per  cent  are  given. 

^(i^-l)  ^ 

,,S==^^S^-^^^PrX     ^ 


What  annuity  for  5  years  will  $4675  give  when  interest 
is  reckoned  at  4  per  cent  ?  ' 

;Sf=  Pr  X  -^—  =  $4675  X  0.04  X  -1^^  =  $1050. 
i2«-l  1.045-1 

373.  Life  Insurance.  In  order  that  a  certain  sum  may  be 
secured,  to  be  payable  at  the  death  of  a  person,  he  pays 
yearly  a  fixed  premiuTYi. 

If  P  denote  the  premium  to  be  paid  for  n  years  to  insure 
an  amount  ^,  to  be  paid  immediately  after  the  last  pre- 
mium, then 

j^^PiR^JIL^.  §368 

R-\ 

.     -p^A{R-Y)_     Ar 


Jin_l  i^«  _  1 

If  A  is  to  be  paid  a  year  after  the  last  premium,  then 
T,_^A(B-1)__        Ar 
R{R--l)      R(R^-l) 

Note.  In  the  calculation  of  life  insurances  it  is  necessary  to  em- 
ploy tables  which  show  for  any  age  the  probable  duration  of  life. 

374.  Bonds.  If  P  denote  the  price  of  a  bond  that  has  7i 
years  to  run,  and  bears  r  per  cent  interest,  S  the  face  of 
the  bond,  and  q  the  current  rate  of  interest,  what  interest 
on  his  investment  will  a  purchaser  of  such  a  bond  receive  ? 

Let  X  denote  the  rate  of  interest  on  the  investment. 

Then  P(l  -f-  xy  is  the  value  of  the  purchase  money  at 
the  end  of  n  years. 


INTEREST   AND   ANNUITIES.  353 

Sr{l  +  qy-''  +  Sr(l  +  qy-'-\- + /S'r  +  ^  is  the  amount 

of  money  received  on  the  bond  if  the  interest  received  from 
the  bond  is  put  immediately  at  compound  interest  at  q  per 
cent. 

But      Sr(l  +  qf-'  +  Sr(l  +  q)--' + +  Sr-{-S 

Sr\(l  +  qY-l'] 


i+"(f+™i±f^)= 


( 


8(/  +  Sr(l  +  gf-Sr 
Pi 


(1)  What  interest  will  a  person  receive  on  his  invest- 
ment if  he  buys  at  114  a  4  per  cent  bond  that  has  26  years 
to  run,  money  being  worth  Z\  per  cent? 


1  +a;  = 


/3.5  +  4(1.035)'^«-4yV 

I,        3.99        ; 


By  logarithms,  1  +  a;  =  1.033. 

That  is,  the  purchaser  will  receive  3J  per  cent  for  his  money. 

(2)  At  what  price  must  7  per  cent  bonds,  running  12 
years,  with  the  interest  payable  semi-annually,  be  bought, 
in  order  that  the  purchaser  may  receive  on  his  investment 
5  per  cent,  interest  semi-annually,  which  is  the  current  rate 
of  interest  ? 

P(l  +  ^Y  =  Sq  +  Sr(l+q)n-8r 

•    r     Sq  +  Sr(l+q)    -  Sr 
5(l+a;)« 
In  this  case  S=-- 100;  and,  as  the  interest  is  serai-annual, 

^==0.025,  r  =  0.035,  n  =  24,  re  =  0.025. 
TT  p     2.5 +  3.5(1.025)'^ -3.5 

^^^^''  ^=         0.025(1.025)- 

By  logarithms,  P=118. 


354  ALGEBRA. 


Exercise  120. 


1.  In  how  many  years  will  $100  amount  to  $1050,  at  5 

per  cent  compound  interest  ? 

2.  In  how  many  years  will  $i4  amount  to  |^  (1)  at 

simple  interest,  (2)  at  compound  interest,  r  and  R 
being  used  in  their  usual  sense  ? 

3.  Find  the  difference  (to  five  places  of  decimals)  be- 

tween the  amount  of  $1  in  2  years,  at  6  per  cent 
compound  interest,  according  as  the-  interest  is  due 
yearly  or  monthly. 

4.  At  5  per  cent,  find  the  amount  of  an  annuity  of  $^ 

which  has  been  left  unpaid  for  4  years. 

5.  Find  the  present  value  of  an  annuity  of  $100  for  5 

years,  reckoning  interest  at  4  per  cent. 

6.  A  perpetual  annuity  of  $1000  is  to  be  purchased,  to 

begin  at  the  end  of  10  years.  If  interest  is  reckoned 
at  3|-  per  cent,  what  should  be  paid  for  it  ? 

7.  A  debt  of  $1850  is  discharged  by  two  payments  of 

$1000  each,  at  the  end  of  one  and  two  years.  Find 
the  rate  of  interest  paid. 

8.  Reckoning  interest  at  4  per  cent,  what  annual  pre- 

mium should  b6  paid  for  30  years,  in  order  to  secure 
$2000  to  be  paid  at  the  end  of  that  time,  the  pre- 
mium being  due  at  the  beginning  of  each  year  ? 

9.  An  annual  premium  of  $  150  is  paid  to  a  life-insurance 

company  for  insuring  $5000.  If  money  is  worth  4 
per  cent,  for  how  many  years  must  the  premium  be 
paid  in  order  that  the  company  may  sustain  no  loss? 


INTEREST   AND    ANNUITIES.  355 

10.  What  may  be  paid  for  bonds  due  in  10  years,  and 

bearing  semi-annual  coupons  of  4  per  cent  each,  in 
order  to  realize  3  per  cent  semi-annually,  if  money 
is  worth  3  per  cent  semi-annually  ? 

11.  "When  money  is  worth  2  per  cent  semi-annually,  if 

bonds  having  12  years  to  run,  and  bearing  semi- 
annual coupons  of  3|-  per  cent  each,  are  bought  at 
114-|-,  what  per  cent  is  realized  on  the  investment? 

12.  If  $126  is  paid  for  bonds  due  in  12  years,  and  yield- 

ing 3|-  per  cent  semi-annually,  what  per  cent  is 
realized  on  the  investment,  provided  money  is  worth 
2  per  cent  semi-annually  ? 

13.  A  person  borrows  $600.25.     How  much  must  he  pay 

annually  that  the  whole  debt  may  be  discharged  in 
35  years,  allowing  simple  interest  at  4  per  cent  ? 

14.  A  perpetual  annuity  of.  f  100  a  year  is  sold  for  $2500. 

At  what  rate  is  the  interest  reckoned  ? 

15.  A  perpetual  annuity  of  $320,  to  begin  10  years  hence, 

is  to  be  purchased.  If  interest  is  reckoned  at  3J- 
per  cejit,  what'  should  be  paid  for  it  ? 

16.  A  sum  of  $10,000  is  loaned  at  4  per  cent.     At  the 

end  of  the  first  year  a  payment  of  $400  is  made; 
and  at  the  end  of  each  following  year  a  payment  is 
made  greater  by  30  per  cent  than  the  preceding 
I  payment.  Find  in  how  many  years  the  debt  will 
be  paid. 

17.  A  man  with  a  capital  of  $100,000  spends  every  year 

$9000.  If  the  current  rate  of  interest  is  5  per  cent, 
in  how  many  years  will  he  be  ruined  ? 

18.  Find  the  amount  of  $365  at  compound  interest  for  20 

years,  at  5  per  cent. 


CHAPTER  XXVIII. 

CHOICE. 

375.  Fundamental  Principle.  If  one  thing  can  he  done  in 
a  different  ways,  and,  when  it  has  been  done,  a  second  thing 
can  he  done  in  b  different  ways,  then  the  two  things  can  he 
done  together  m  a  X  b  different  ways. 

For,  corresponding  to  the  first  way  of  doing  the  first 
thing,  there  are  h  different  ways  of  doing  the  second  thing ; 
corresponding  to  the  second  way  of  doing  the  first  thing, 
there  are  h  different  ways  of  doing  the  second  thing ;  and 
so  on  for  each  of  the  a  dijQTerent  ways  of  doing  the  first 
thing.  Therefore  there  are  aXh  different  ways  of  doing 
the  two  things  together. 

(1)  If  a  box  contains  four  capital  letters.  A,  B,  C,  D, 
and  three  small  letters,  x,  y,  z,  in  how  many  different  ways 
may  two  letters,  one  a  capital  letter  and  one  a  small  letter, 
be  selected  ? 

A  capital  letter  may  be  selected  in  four  different  ways,  since  any 
one  of  the  letters  A,  B,  C,  D,  may  be  selected.  A  small  letter  may 
be  selected  in  three  different  ways,  since  any  one  of  the  letters  x,  y,  2, 
may  be  selected.  Any  small  letter  may  be  put  with  any  capital 
letter. 

Thus,  with  A  we  may  put  x,  or  3/,  or  2 ; 

with  B  we  may  put  x,  or  y,  or  2 ; 

with  C  we  may  put  x,  or  3/,  or  2 ; 

with  D  we  may  put  x,  or  y,  or  2. 

Hence  the  number  of  ways  in  which  a  selection  may  be  made  is 

4  X  3,  or  12.     These  ways  are  : 

Ax  Bx  Cx  Dx 

Ay  By  Cy  Dy 

Az  Bz  Cz  Dz 


CHOICE.  357 

(2)  On  a  shelf  are  7  English,  5  French,  and  9  German 
books.  In  how  many  ways  may  two  books,  not  in  the 
same  language,  be  selected  ? 

An  English  book  and  a  French  book  can  be  selected  in  7  X  5,  or 
35,  ways.  A  French  book  and  a  German  book  in  5  X  9,  or  45,  ways. 
An  English  book  and  a  German  book  in  7  X  9,  or  63,  ways. 

Hence,  there  is  a  choice  of  35  +  45  +  63,  or  143,  ways. 

(3)  Out  of  the  ten  figures,  0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  how 
many  numbers,  each  consisting  of  two  figures,  can  be  formed? 

Since  0  has  no  value  in  the  left-hand  place,  the  left-hand  place 
can  be  filled  in  9  ways. 

The  right-hand  place  can  be  filled  in  10  ways,  since  repetitions  of 
the  digits  are  allowed  (as  22,  33,  etc.). 

Hence,  the  whole  number  is  9  X  10,  or  90. 

376.  By  successive  application  of  the  principle  of  §  375 
it  may  be  shown  that, 

If  one  thing  can  be  done  in  a  different  ivays,-  and  then  a 
second  thing  can  be  done  m  b  different  ways,  then  a  third 
thing  in  c  different  ways,  then  a  fourth  thing  in  d  diffei^ent 
ways,  etc.,  the  number  of  different  ways  of  doing  all  the 
things  together  will  5e  a  X  b  X  c  X  d,  etc. 

For,  the  first  and  second  things  can  be  done  together  in 
axb  different  ways  (§  375),  and  the  third  thing  in  c  differ- 
ent ways ;  hence,  (§  375),  the  first  and  second  things  and 
the  third  thing  can  be  done  together  m  (axb)x  c  differ- 
ent ways.  Therefore,  the  first  three  things  can  be  done  in 
axbx  G  different  ways.  And  so  on  for  any  number  of 
things. 

In  how  many  ways  can  four  Christmas  presents  be  given 
to  four  boys,  one  to  each  boy  ? 

The  first  present  may  be  given  to  any  one  of  the  boys ;  hence 
there  are  4  ways  of  disposing  of  it. 


358  ALGEBRA. 

The  second  present  may  be  given  to  any  one  of  the  other  three 
boys ;  hence  there  are  3  ways  of  disposing  of  it. 

The  third  present  may  be  given  to  either  of  the  other  two  boys ; 
hence  there  are  2  ways  of  disposing  of  it. 

The  fourth  present  must  be  given  to  the  last  boy ;  hence  there  is 
only  1  way  of  disposing  of  it. 

There  are,  then,  4  x  3  x  2  X  1,  or  24,  ways. 

377.   Selections  and  Arrangements. 

(1)  In  how  many  ways  can  a  vowel  and  a  consonant  be 
chosen  out  of  the  alphabet  ? 

Since  there  are  in  the  alphabet  6  vowels  and  20  consonants,  a 
vowel  can  be  chosen  in  6  ways  and  a  consonant  in  20  ways,  and 
both  (§  375)  in  6  X  20,  or  120,  ways. 

(2)  In  how  many  ways  can  a  two-lettered  word  be 
made,  containing  one  vowel  and  one  consonant  ? 

The  vowel  can  be  chosen  in  6  ways  and  the  consonant  in  20 
ways ;  and  then  each  combination  of  a  vowel  and  a  consonant  can 
be  written  in  2  ways ;  as  ac,  ca. 

Hence,  the  whole  number  of  ways  is  6  X  20  x  2,  or  240. 

These  two  examples  show  the  difference  between  a  selec- 
tion or  combination  of  different  things,  and  an  arrangement 
or  permutation  of  the  same  things. 

Thus,  ac  forms  a  selection  of  a  vowel  and  a  consonant,  and  ac  and 
ca  form  two  different  arrangements  of  this  selection. 

From  (1)  it  is  seen  that  120  different  selections  can  be  made  with 
a  vowel  and  a  consonant ;  and  from  (2)  it  is  seen  that  240  different 
arrangements  can  be  made  with  the  same. 

Again,  a,  &,  c  is  a  selection  of  three  letters  from  the  alphabet. 
This  selection  admits  of  6  different  arrangements,  as  follows : 
abc  bea  cab 

acb  bac  cba 

A  selection  or  combination  of  any  number  of  things  is  a 
group  of  that  number  of  things  put  together  without  regard 
to  their  order. 


CHOICE.  359 

An  arrangement  or  permutation  of  any  number  of  things  is 
a  group  of  that  number  of  things  put  together,  regard  being 
paid  to  their  order. 

378.  Arrangements,  Things  all  Different.  The  number  of 
different  arrange'tnents  {or  'permutations)  of  n  different  things 
taken  all  together  is 

n{n  -  l){n  —  2)(n  -  8) 3  X  2  X  1. 

For,  the  first  place  can  be  filled  in  n  ways,  then  the 
second  place  in  n  —  1  ways,  then  the  third  place  in  7i  —  2 
ways,  and  so  on  to  the  last  place,  which  can  be  filled  in 
only  1  way. 

Hence  (§  376)  the  whole  number  of  arrangements  is  the 
continued  product  of  all  these  numbers, 

n(n  -  l)(n  -  2)(n  -  3) 3x2x1. 

For  the  sake  of  brevity  this  product  is  written  \n,  and  is 
read  factorial  n. 

Observe  that  1x2 (n  —  l)n^\n. 

How  many  different  arrangements  of  nine  letters  each 
can  be  formed  with  the  letters  in  Cambridge  ? 

There  are  nine  letters.  In  making  any  arrangement  any  one  of 
the  letters  can  be  put  in  the  first  place.  Hence,  the  first  place  can 
be  filled  in  9  ways. 

Then  the  second  place  can  be  filled  with  any  one  of  the  remain- 
ing eight  letters ;  that  is,  in  8  ways. 

In  like  manner,  the  third  place  can  be  filled  in  7  ways,  the  fourth 
place  in  6  ways,  and  so  on ;  and,  lastly,  the  ninth  place  in  1  way. 

If  the  nine  places  are  indicated  by  Roman  numerals,  ^e  result 
is  {I  376)  as  follows  : 

I.   II.  III.  IV.  V.  VI.  VII.  VIII.  IX. 
9x8x7x6x5x4x    3    x  2  x  1  =  362,880  ways. 

Hence,  there  are  362,880  different  arrangements  possible. 


360  ALGEBRA. 

379.  The  number  of  different  arrangements  of  n  different 
things  taken  i  at  a  time  is 

n{n  —  V){n  —  2) to  r  factors, 

that  is,  n(n  —  l)(n  —  2) [n  —  (r  —  1)], 

or  n{n—l)(n  —  2) (72  — r  4-1). 

For,  the  first  place  can  be  filled  in  n  ways,  the  second 
place  in  71  —  1  ways,  the  third  place  in  n  —  2  ways,  and  the 
rth  place  in  n  —  (r—l)  ways. 

Let  Pn^r  represent  the  number  of  arrangements  of  n  dif- 
ferent things  taken  r  at  a  time.     Then  • 

-Pn,r  =  '^(^  —  1)(^  —  2) to  r  factors 

=  n(n~l)(n  —  2) (n  —  r-}-l). 

How  many  different  arrangements  of  four  letters  each 
can  be  formed  from  the  letters  in  Cambridge^ 
There  are  nine  letters  and  four  places  to  be  filled. 
The  first  place  can  be  filled  in  9  ways.     Then  the  second  place 
can  be  filled  in  8  ways.     Then  the  third  place  in  7  ways,  and  the 
fourth  place  in  6  ways. 

If  the  places  are  indicated  by  I.,  II.,  III.,  IV.,  the  result  is  (^  376) 
I.    II.  III.  IV. 
9x8x7x6  =  3024  ways. 

Hence,  there  are  3024  different  arrangements  possible. 

380.  Selections,  Things  all  Different.  The  number  of  df- 
ferent  selections  (or  combinatiojis)  of  n  different  things  taken 
X  at  a  time  is 

n(n-  l)(n  —  2) (n  —  r-j-l) 

\L 
To  prove  this,  let  Cn,r  represent  the  number  of  different 
selections  (or  combinations)  of  n  different  things  taken  r  at 
a  time. 

Take  one  selection  of  r  things;  from  this  selection  [r 
arrangements  can  be  made  (§  378). 


CHOICE.  361 

Take  a  second  selection ;  from  this  selection  [r  arrange- 
ments can  be  made.  And  so  on  for  each  of  the  (7„_^  selec- 
tions. 

Hence,  (7„, ,.  X  [r  is  the  number  of  arrangements  of  n  dif- 
ferent things  taken  r  at  a  time  ;  or 

\r 

.    n     _n{n-l)(7i-2) (n-r-\-l) 

\r 
Ex.    In  how  many  different  ways  can  three  vowels  be 
selected  from  the  five  vowels  a,  e,  i,  o,  u? 

The  number  of  different  ways  in  which  we  can  arrange  3  vowels 
out  of  5  is  (g  376)  5  X  4  X  3,  or  60. 

These  60  arrangements  might  be  obtained  by  first  forming  all  the 
possible  selections  of  3  vowels  out  of  5,  and  then  arranging  the  3 
vowels  in  each  selection  in  as  many  ways  as  possible. 

Since  each  selection  can  be  arranged  in  13,  or  6  ways  (|  378),  the 
number  of  selections  is  -^^  or  10. 

The  formula  applied  to  this  problem  gives 
Q.3^5X4X3^ 
^' '      1  2<  2  X  3 

381.   Selections,  Second  Pormula.    Multiplying  both  numer- 
ator and  denominator  of  the  expression  for  the  number  of 
selections  in  the  last  example  by  2  X  1,  we  have 
C    -^X4x  3x2x  1_   [5 
'•'      1x2x3x2x1"  [3|2' 
In  general,  multiplying  both  numerator  and  denomina- 
tor of  the  expression  for  (7„, / in  §  380  by  \n  —  r,  we  have 

C     —  ^(^  ~  1) (^  —  '^+  l)(w  — r) 1 

"■'■""  \7^X(n-r) 1 

\r\n  —  r 


362  ALGEBRA. 

This  second  form  is  more  compact  than  the  first,  and  is 
more  easily  remembered. 

Note.   In  reducing  a  result  expressed  in  the  above  form,  it  is  to 

be  observed  that  \n  —  r  cancels  all  the  factors  of  the  numerator  from 

112 
1  up  to  and  including  n  —  r.     Thus,  in  ■^=^,  [7  cancels  all  the  fac- 

[5  [7 

tors  of  [12  from  1  up  to  and  including  7 ;  so  that 

Uj  _  12x11x10x9x8  _  ^g2 
[5[7         1x2x3x4x5 

382.  Theorem.  The  number  of  selections  ofn  things  taken 
Y  at  a  time  is  the  same  as  the  number  of  selections  of  n  things 
taken  n—  i  at  a  time. 

For     C  1-^  __      1-^       _rv 

\n  —  r\n—  {n  —  r)      \n  —  r\r 

This  is  also  evident  from  the  fact  that  for  every  selection 
of  r  things  taJcen,  a  selection  of  n  —  r  things  is  left. 

Thus,  out  of  8  things,  3  things  can  be  selected  in  the  same  number 
of  ways  as  5  things ;  namely, 

Ji.^  8X7X6  ^53 
1^15  [3 

|1         1 
Note.   Evidently  d,  i  =  1 ;  also  Ci.  i  =  -fj-  =  — • 

.•.i  =  l.  and  10  =  1. 

|0       '  L 

383.  Examples  in  Selections   and   Arrangements.      Of  the 

arrangements  possible  with  the  letters  of  the  word  Cam- 
bridge, taken  all  together, 

(1)   How  many  will  begin  with  a  vowel? 

In  filling  the  nine  places  of  any  arrangement  the  first  place  can 
be  filled  in  only  3  ways,  the  other  places  in  [8  ways. 
Hence,  the  answer  is  (§  376) 

3  X  [8  =  120,960. 


CHOICE.  363 

(2)  How  many  will  both  begin  and  end  with  a  vowel  ? 

The  first  place  can  'be  filled  in  3  ways,  the  last  place  in  2  ways 
(one  vowel  having  been  used),  and  the  remaining  seven  places  in 
[7  ways. 

Hence,  the  answer  is  (^  376) 

3x2x17  =  30,240. 

(3)  How  many  will  begin  with  Cam  ? 

The  answer  is  evidently  [6 ;  since  our  only  choice  lies  in  arrang- 
ing the  remaining  six  letters  of  the  word. 

(4)  How  many  will  have  the  letters  cam  standing 
together  ? 

This  may  be  resolved  into  arranging  the  group  cam  and  the  last 
six  letters,  regarded  as  seven  distinct  elements,  and  then  arranging 
the  letters  cam. 

The  first  can  be  done  in  [7  ways,  and  the  second  in  [3  ways. 
Hence  both  can  be  done  in  [7  X  [3  =  30,240  ways. 

In  how  many  ways  can  the  letters  of  the  word  Cam- 
bridge be  written, 

(5)  Without  changing  ih.Q place  of  any  vowel? 

The  second,  sixth,  and  ninth  places  can  be  filled  each  in  only  1 
way  ;  the  other  places  in  [6  ways. 

Therefore,  the  whole  number  of  ways  is  1_6  =  720. 

(6)  Without  changing  the  order  of  the  three  vowels  ? 

The  vowels  in  the  different  arrangements  are  to  be  kept  in  the 
order,  a,  t,  e. 

One  of  the  six  consonants  can  be  placed  in  4  ways :  before  a,  be- 
tween a  and  i,  between  i  and  e,  and  after  e. 

Then  a  second  consonant  can  be  placed  in  5  ways,  a  third  conso- 
nant in  6  ways,  a  fourth  consonant  in  7  ways,  a  fifth  consonant  in  8 
ways,  and  the  last  consonant  in  9  ways.  Hence  the  whole  number 
of  ways  is 

4x5x6x7x8x9,  or  60,480. 


364  ALGEBRA. 

(7)  Out  of  20  consonants,  in  how  many  ways  can  18  be 
selected  ? 

The  18  can  be  selected  in  the  same  number  of  ways  as  2 ;  and  the 
number  of  ways  in  which  2  can  be  selected  is 

20X29^190. 
2 

(8)  In  how  many  ways  can  the  same  choice  be  made  so 
as  always  to  include  the  letter  h  ? 

Taking  h  first,  we  must  then  select  17  out  of  the  remaining  19 
consonants.     This  can  be  done  in 

li|i»  =  171  ways. 

(9)  In  how  many  ways  can  the  same  choice  be  made  so 
as  to  include  b  and  not  to  include  c? 

Taking  h  first,  we  have  then  to  choose  17  out  of  18,  c  being  ex- 
cluded.    This  can  be  done  in  18  ways. 

(10)  From  20  Kepublicans  and  6  Democrats,  in  how 
many  ways  can  5  different  offices  be  filled,  3  of  which 
must  be  filled  by  Republicans,  and  the  other  2  by 
Democrats  ? 

The  first  three  offices  can  be  assigned  to  3  Republicans  in 
20  X  19  X  18  -  6840  ways ; 
and  the  other  two  offices  can  be  assigned  to  2  Democrats  in 
6  X  5  =  30  ways. 
There  is,  then,  a  choice  of  6840  X  30  =  205,200  difl^erent  ways. 

(11)  Out  of  20  consonants  and  6  vowels,  in  how  many 
ways  can  we  make  a  word  consisting  of  3  different  conso- 
nants and  2  different  vowels  ? 

20  X  19  X  18 

Three  consonants  can  be  selected  in  — — —  =  1140  ways, 

6x5  1X2X3 

and  two  vowels  in  =  15  ways.     Hence  the  5  letters  can   be 

1x2  ^ 

selected  in  1140  x  15  =  17,100  ways. 


CHOICE.  365 

When  five  letters  have  been  so  selected,  they  can  be  arranged  in 
|5_=  120  different  orders.  Hence,  there  are  17,100  X  120  =  2,052,000 
different  ways  of  making  the  word. 

Observe  that  the  letters  are  first  selected  and  then  arranged. 

(12)  A  society  consists  of  50  members,  10  of  whom  are 
physicians.  In  how  many  ways  can  a  committee  of  6 
members  be  selected  so  as  to  include  at  least  one  physician  ? 

Six  members  can  be  selected  from  the  whole  society  m 
(50 

Six  members  can  be  selected  from  the  whole  society,  so  as  to  in- 
clude no  physician,  by  choosing  them  all  from  the  40  members  who 
are  not  physicians,  and  this  can  be  done  in 

140 

. ways. 

I^[34       ^ 

150         [40 
Hence,  —  is  the  number  of  ways  of  selecting 

16 [44     [6 1 34 

the  committee  so  as  to  include  at  least  one  physician. 

384.   Greatest  Number   of   Selections,     To   find  for  what 
value  of  r  the  number  of  selections  of  n  things,  taken  r  at 
a  time,  is  the  greatest. 
The  formula 

^     _n{n-  V){n  —  2) {n-r-\- 1) 

Ix2x3x r 

may  be  written 

^    nn  —  Yn  —  ^       n  —  r-\-\ 

6.,,  -.  X  ^-X  -^ 

The  numerators  of  the  factors  on  the  right  side  of  this 
equation  begin  with  n,  and  form  a  descending  series  with 
the  common  difference  1 ;  and  the  denominators  begin  with 
1,  and  form  an  ascending  series  with  the  common  difference 
1.     Therefore,  from  some  point  in  the  series,  these  factors 


366  ALGEBRA. 

become  less  than  1.  Hence,  the  maximum  product  is 
reached  when  that  product  includes  all  the  factors  greater 
than  1. 

I.  When  n  is  an  odd  number,  the  numerator  and  the 
denominator  of  each  factor  will  be  alternately  both  odd  and 
both  even ;  so  that  the  factor  greater  than  1,  but  nearest 
to  1,  will  be  the  factor  whose  numerator  exceeds  the  de- 
nominator by  2.  Hence,  in  this  case,  r  must  have  such  a 
value  that 

n  —  r  -\-l  =  r-\-2,  or  r  =  — - — 

II.  When  n  is  an  even  number,  the  numerator  of  the  first 
factor  will  be  even  and  the  denominator  odd ;  the  numer- 
ator of  the  second  factor  will  be  odd  and  the  denominator 
even ;  and  so  on,  alternately ;  so  that  the  factor  greater 
than  1,  but  nearest  to  1,  will  be  the  factor  whose  numerator 
exceeds  the  denominator  by  1.  Hence,  in  this  case,  r  must 
have  such  a  value  that 

n  —  r-\-l  =  r  -\-l,  or  ?"  — -• 

(1)  What  value  of  r  will  give  the  greatest  number  of 
selections  out  of  7  things  ? 

Here  n  is  odd,  and  r  =  — — -  =  — —  =  3. 

.^^^7x6x5^3g 
1x2x3 

Ifr  =  4.then  ,^7x6x5x4^3^ 

1x2x3x4 

When  the  number  of  things  is  odd,  there  will  be  two  equal  num- 
bers of  selections  ;  namely,  when  the  number  of  things  taken  together 
is^'wsi  under  and  jwsi  over  one-half  of  the  whole  number  of  things. 


CHOICE.  367 

(2)  What  value  of  r  will  give  the  greatest  number  of 
selections  out  of  8  things  ? 


2  2 

8x7x6x5 

1x2x3x4 


70. 


So  that,  when  the  number  of  things  is  even,  the  number  of  selec- 
tions will  be  greatest  when  one-half  of  the  whole  are  taken  together. 

385,  Division  into  Groups.  The  number  of  diiferent  ways 
in  which  p-\-  q  things  all  different  can  be  divided  into  two 
groups  of  p  things  and  q  things,  respectively,  is  the  same 
as  the  number  of  ways  in  which  p  things  can  be  selected 

from  p-{-  q  things,  or  ^      ^. 

For,  to  each  selection  of  p  things  taken  corresponds  a 
selection  of  q  things  left,  and  each  selection  therefore  effects 
the  division  into  the  required  groups. 

(1)  In  how  many  ways  can  18  men  be  divided  into  2 
groups  of  6  and  12  each  ? 

[6L12 

(2)  A  boat's  crew  consists  of  8  men,  of  whom  2  can  row 
only  on  the  stroke  side  of  the  boat,  and  3  can  row  only 
on  the  bow  side.  In  how  many  ways  can  the  crew  be 
arranged  ? 

There  are  left  3  men  who  can  row  on  either  side ;  2  of  these  must 
row  on  the  stroke  side,  and  1  on  the  bow  side. 

The  number  of  ways  in  which  these  three  can  be  divided  is 

_  =  3ways. 


368  ALGEBRA. 

When  the  stroke  side  is  completed,  the  4  men  can  be  arranged 
in  |_4  ways ;  likewise,  the  4  men  of  the  bow  side  can  be  arranged 
in  |_4  ways.     Hence  the  arrangement  can  be  made  in 
3  X  |4  X  [£=  1728  ways. 

386.  The  number  of  different  ways  in  which  p  -{-  q-}-0' 
things  all  different  can  be  divided  into  three  groups  of  p 

things,  q  things,  and  r  things,  respectively,  is  Ir  ■"  2^  ~^- 

For,  p-}-  q  +  r  things  may  be  divided  into  two  groups 

\p-i-q  +  r 
of  p  things  and  g'  +  r  things  in  -. — j — -- —  ways  ;  then,  the 

group  of  q-\-r  things  may  be  divided  into  two  groups  of 

\q  +  r 
q  things  and  r  things  in  —. — j —  ways ;  hence  the  division 

into  three  groups  may  be  effected  in 

\p  +  q  +  r      \q  +  r       \p-hq  +  r 
\p\9  +  r         l^|r    °'      l^illr      ^^^^^* 
And  so  on  for  any  number  of  groups. 

In  how  many  ways  can  a  company  of  100  soldiers  be 

divided  into  three  squads  of  50,  30,  and  20,  respectively  ? 

1 100 
The  answer  is  ^^^  ways. 

387.  When  the  number  of  things  is  the  same  in  two  or 
more  groups,  and  there  is  no  distinction  to  be  made  between 
these  groups,  the  number  of  ways  given  by  the  preceding 
section  is  too  large. 

Divide  the  letters  a,  b,  c,  d,  into  two  groups  of  two  let- 
ters each. 

[4 
The  number  of  ways  given  by  §  386  is  -^j^  =  6 ;  these  ways  are : 


CHOICE.  369 

I.  ab  cd.  IV.   be  ad. 

II.   ac   bd.  V.   bd  ac. 

III.  ad  be.  VI.   cd  ab. 

Since  there  is  no  distinction  between  the  groups,  group  IV.  is  the 
same  as  group  III.,  group  V.  the  same  as  group  II.,  and  group  VI. 

the  same  as  group  I.     Hence,  the  correct  answer  is  -  X  -f^'  ^^  ^• 

l_~  _ 

In  the  case  of  three  similar  groups  the  result  given  by 
§  386  is  to  be  divided  by  1 3^,  the  number  of  ways  in  which 
three  groups  can  be  arranged  among  themselves ;  in  the 
case  of  four  groups,  by  [4_;  and  so  on. 

In  how  many  ways  can  18  men  be  divided  into  2  groups 

of  9  each? 

[18 
According  to  ^  386,  the  answer  would  be  -— — • 

|_9[9 

The  two  groups,  considered  as  groups,  have  no  distinction ; 
therefore,  permuting  them  gives  no  new  arrangement,  and  the  true 

|18 
result  is  obtained  by  dividing  the  preceding  by  [2,  and  is     ~~'    . 

[2|_9[9 

If  any  condition  be  added  that  will  make  the  two  groups  different, 
if,  for  example,  one  group  wear  red  badges  and  the  other  blue,  then 

118 

the  answer  will  be  -——• 

388.  Arrangements,  Kepetitions  allowed.  Suppose  we  have 
n  different  letters,  and  that  repetitions  are  allowed. 

Then,  in  making  any  arrangement,  the  first  place  can  be 
filled  in  n  ways ;  and  the  second  place  can  be  filled  in  n 
ways,  since  repetitions  are  allowed.  Hence  the  first  two 
places  can  be  filled  in  n  X  w,  or  n^,  ways  (§  375). 

Similarly,  the  first  three  places  can  be  filled  in  nXnXn, 
or  n',  ways  (§  376). 

In  general,  r  places  can  be  filled  in  rf  ways ;  or,  the 
number  of  arrangements  of  n  different  things  taken  i  at  a 
time,  when  repetitions  are  allowed,  is  n'". 


370  ALGEBRA. 

(1)  How  many  three-lettered  words  can  be  made  from 
the  alphabet,  when  repetitions  are  allowed  ? 

Here  the  first  place  can  be  filled  in  26  ways ;  the  second  place  in 
26  ways ;  and  the  third  place  in  26  ways.  The  number  of  words  is, 
therefore,  26'  =  17,576. 

(2)  In  the  common  system  of  notation,  how  many  num- 
bers can  be  formed,  each  number  consisting  of  not  more 
than  5  figures? 

Each  of  the  possible  numbers  may  be  regarded  as  consisting  of 
5  figures,  by  prefixing  zeros  to  the  numbers  consisting  of  less  than  5 
figures.     Thus,  247  may  be  written  00247. 

Hence,  every  possible  arrangement  of  5  figures  out  of  the  10 
figures,  except  00000,  will  give  one  of  the  required  numbers ;  and 
the  answer  is  10^  —  1  =  99,999 ;  that  is,  all  the  numbers  between  0 
and  100,000. 


389.  Arrangements,  Things  Alike,  All  together.  Consider 
the  number  of  arrangements  of  the  letters  a,  a,  b,  b,  b,  c,  d. 

Suppose  the  a's  to  be  different  and  the  &'s  to  be  different,  and  dis- 
tinguish between  them  by  a^,  a^,  b^,  b^,  b^. 

The  seven  letters  can  now  be  arranged  in  [7  ways  (g  376). 

Now  suppose  the  two  a's  to  become  alike,  and  the  three  6's  to  be- 
come alike.  Then,  where  we  before  had  |_2  arrangements  of  the  a's 
among  themselves,  we  now  have  but  one  arrangement,  aa;  and 
where  we  before  had  [3  arrangements  of  the  6's  among  themselves, 
we  now  have  but  one  arrangement,  bbb. 

[7 

Hence,  the  number  of  arrangements  is  -— —  =  420. 

Xn  general,  the  number  of  arrangements  of  n  things,  of 
which  p  are  alike,  q  others  are  alike,  and  r  others  are  alike, 
,  is 


V| 


CHOICE.  371 

(1)  In  how  many  ways  can  the  letters  of  the  word  Col- 
lege be  arranged  ? 

If  the  two  Ts  were  different  and  the  two  e's  were  different,  the 
number  of  ways  would  be  [7.  Instead  of  two  arrangements  of  the 
two  Z's,  we  have  but  one  arrangement,  II  \  and  instead  of  two  ar- 
rangements of  the  two  e's,  we  have  but  one  arrangement,  ee.    Hence, 

\1 
the  number  of  ways  is  — — -  =  1260. 
^        [2[2 

(2)  In  how  many  different  orders  can  a  row  of  4  white 
bails  and  3  black  balls  be  arranged? 

[4  [3 

390.  Selections,  Eepetitions  allowed.  We  will  illustrate 
by  an  example  the  method  of  solving  problems  that  come 
under  this  head. 

In  how  many  ways  can  a  selection  of  3  letters  be  made 
from  the  letters  a,  b,  c,  d,  e,  if  repetitions  are  allowed? 
The  selections  will  be  of  three  classes : 

(a)  All  three  letters  alike. 
(6)  Two  letters  alike, 
(c)  The  three  letters  all  different. 
(a)  There  will  be  5  selections,  since  any  one  of  the  five  letters 
may  be  taken  three  times. 

(6)  Any  one  of  the  five  letters  may  be  taken  twice,  and  with  these 
may  be  put  any  one  of  the  other  four  letters.  Hence,  the  number 
of  selections  is  5  X  4,  or  20. 

(c)  The  number  of  selections  (g  380)  is  '^  ^  ^  ^  |  or  10.     Hence, 

the  total  number  of  selections  is  5  +  20  +  10  =  35o 

391.  Selections  and  Arrangements,  Things  AHke.  We  will 
illustrate  by  an  example  the  method  of  solving  problems 
that  come  under  this  head. 

How  many  selections  of  four  letters  each  can  be  made 
from  the  letters  in  Proportion  ?  How  many  arrangements 
of  four  letters  each  ? 


372  ALGEBRA. 

There  are  10  letters  as  follows : 

0      p      r      t     i      n 
0      p      r 

0 

Selections : 

The  selections  may  be  divided  into  four  classes : 

(a)  Three  letters  alike. 

(6)  Two  letters  alike,  two  others  alike. 

(c)  Two  letters  alike,  other  two  different. 

(d)  Four  letters  different. 

(a)  With  the  three  o's  we  may  put  any  one  of  the  five  other  letters, 
giving  5  selections. 

(6)  We  may  choose  any  two  out  of  the  three  pairs,  o,o;  p,p;  r,  r. 

^     =  3  selections. 
1X2 

(c)  With  any  one  of  the  three  pairs  we  can  put  any  two  of  the 
five  remaining  letters  in  the  first  line. 

3  X  ^^  =  30  selections. 
1x2 

(d)  6x5x4x3  _  ^5  selections. 
^^  1x2x3x4 

Hence,  the  total  number  of  selections  is 

5  +  3  +  30  +  15  =  53. 
Arrangements : 

(a)  Each  selection  can  be  arranged  in  rr  =  4  ways. 

li 
5  X  4  =  20  arrangements. 

[4 
{!))  Each  selection  can  be  arranged  in  -^-  =  6  ways. 

3  X  6  =  18  arrangements. 

\± 
(c)  Each  selection  can  be  arranged  in  —  =  12  ways. 

30  X  12  =  360  arrangements. 
{d)  Each  selection  can  be  arranged  in  [4  =  24  ways. 

15  X  24  =  360  arrangements. 
Hence,  the  total  number  of  arrangements  is 

20  +  18  +  360  +  360  -  758. 


CHOICE.  373 

392.   Total  Number  of  Selections. 

I.  The  whole  number  of  ways  in  which  a  selection  (of 
some,  or  all)  can  be  made  from  n  different  things  is  2"  —  1. 

For  each  thing  can  be  either  taken  or  left ;  that  is,  can 
be  disposed  of  in  two  ways.  There  are  n  things;  hence 
(§  376)  they  can  all  be  disposed  of  in  2"  ways.  But  among 
these  ways  is  included  the  case  in  which  all  are  rejected ; 
and  this  case  is  inadmissible.  Hence,  the  number  of  ways  of 
making  a  selection  is  2**  —  1. 

How  many  different  amounts  can  be  weighed  with  1  lb., 
2  lb.,  4  lb.,  8  lb.,  and  16  lb.  weights? 

25  - 1  =  31. 

II.  The  whole  number  of  loays  in  which  a  selection  can 

he  made  from  p  +  q  +  r things,  of  which  p  are  alike,  q 

are  alike,  r  are  alike,  etc.,  is  (p  -{-  l)(q  +  l)(r  -f- 1 —  1). 

For  the  set  of  p  things  may  be  disposed  of  in  ^  +  1 

ways,  since  none  of  them  may  be  taken,  or  1,  2,  3,  , 

or  p,  may  be  taken. 

In  like  manner,  the  q  things  may  be  disposed  of  in  ^  -f-  1 
ways ;  the  r  things  in  r  +  1  ways ;  and  so  on. 

Hence  (§  376)  all  the  things  may  be  disposed  of  in 
(^  +  l)(^  +  l)(r+l) ways. 

But  the  case  in  which  all  the  things  are  rejected  is  in- 
admissible ;  hence,  the  whole  number  of  ways  is 

(p  +  l)(q  +  l)(r+l) -1. 

In  how  many  ways  can  2  boys  divide  between  them 
10  oranges  all  alike,  15  apples  all  alike,  and  20  peaches 
all  alike  ? 

Here,  the  case  in  which  the  first  boy  takes  none,  and  the  case  in 
which  the  second  boy  takes  none,  must  be  rejected. 

Therefore,  the  answer  is  one  less  than  the  result,  according  to  II. 
11  X  16  X  21  -  2  =  3694. 


374  ALGEBRA. 

Exercise   121. 

1.  How  many  different  permutations  can  be  made  of  the 

letters  in  the  word  Ecclesiastical,  taken  all  together  ? 

2.  Of  all  the  numbers  that  can  be  formed  with  four  of  the 

digits  5,  6,  7,  8,  9,  how  many  will  begin  with  56  ? 

3.  If  the  number  of  permutations  of  n  things,  taken  4 

together,  be  equal  to  12  times  the  permutations  of 
n  things,  taken  2  together,  find  n. 

4.  With  3  consonants  and  2  vowels,  how  many  words  of  3 

letters  can  be  formed,  beginning  and  ending  with  a 
consonant,  and  having  a  vowel  for  the  middle  letter? 

5.  Out  of  20  men,  in  how  many  different  ways  can  4  men 

be  chosen  to  be  on  guard  ?  In  how  many  of  these 
would  one  particular  man  be  taken,  and  from  how 
many  would  he  be  left  out  ? 

6.  Of  12  books  of  the  same  size,  a  shelf  will  hold  5.    How 

many  different  arrangements  on  the  shelf  may  be 
made  ? 

7.  Of  8  men  forming  a  boat's  crew,  one  is  selected  as 

stroke.  How  many  arrangements  of  the  rest  are  pos- 
sible? When  the  4  men  who  row  on  each  side  are  de- 
cided on,  how  many  arrangements  are  still  possible  ? 

8.  How  many  signals  may  be  made  with  6  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or 
any  number  at  a  time  ? 

9.  How  many  signals  may  be  made  with  8  flags  of  differ- 

ent colors,  which  can  be  hoisted  either  singly,  or 
any  number  at  a  time  one  above  another  ? 

10.  How  many  different  signals  can  be  made  with  10  flags, 
of  which  3  are  white,  2  red,  and  the  rest  blue, 
always  hoisted  all  together  and  one  above  another? 


CHOICE.  375 

11.  How  many  signals  can  be  made  with  7  flags,  of  which 

2  are  red,  1  white,  3  blue,  and  1  yellow,  always 
displayed  all  together  and  one  above  another  ? 

12.  In  how  many  different  ways  may  the  8  men  serving  a 

field-gun  be  arranged,  so  that  the  same  man  may 
always  lay  the  gun  ? 

13.  Find  the  number  of  signals  which  can  be  made  with  4 

lights  of  different  colors  when  displayed  any  number 
at  a  time,  arranged  one  above  another,  side  by  side, 
or  diagonally. 

14.  From  10  soldiers  and  8  sailors,  how  many  different 

parties  of  3  soldiers  and  3  sailors  can  be  formed  ? 

15.  How  many  signals  can  be  made  with  3  blue  and  2 

white  flags,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  one  above  another  ? 

16.  In  how  many  ways  can  a  party  of  6  take  their  places 

at  a  round  table  ? 

17.  Out  of  12  Democrats  and  16  Republicans,  how  many 

different  committees  can  be  formed,  each  committee 
consisting  of  3  Democrats  and  4  Republicans  ? 

18.  From  12  soldiers  and  8  sailors,  how  many  different 

parties  of  3  soldiers  and  2  sailors  can  be  formed  ? 

19.  Find  the  number  of  combinations  of  100  things,  97 

together. 

20.  With  20  consonants  and  5  vowels,  how  many  different 

words  can  be  formed  consisting  of  3  different  conso- 
nants and  2  different  vowels,  any  arrangement  of 
letters  being  considered  a  word  ? 

21.  Of  30  things,  how  many  must  be  taken  together  in 

order  that,  having  that  number  for  selection,  there 
may  be  the  greatest  possible  variety  of  choice  ? 


376  ALGEBRA. 

22.  There  are  m  things  of  one  kind  and  n  of  another; 

how  many  different  sets  can  be  made  containing  r 
things  of  the  first  and  s  of  the  second  ? 

23.  The  number  of  combinations  of  n  things,  taken  r  to- 

gether, is  3  times  the  number  taken  r  —  1  together, 
and  half  the  number  taken  r  +  1  together.  Find  n 
and  r. 

24.  In  how  many  ways  may  12  things  be  divided  into  3 

sets  of  4  ? 

25.  How  many  words  of  6  letters  may  be  formed  of  3 

vowels  and  3  consonants,  the  vowels  always  having 
the  even  places  ? 

26.  From  a  company  of  90  men,  20  are  detached  for  mount- 

ing guard  each  day.  How  long  will  it  be  before  the 
same  20  men  are  on  guard  together,  supposing  the 
men  to  be  changed  as  much  as  possible ;  and  how 
many  times  will  each  man  have  been  on  guard? 

27.  Supposing  that  a  man  can  place  himself  in  3  distinct 

attitudes,  how  many  signals  can  be  made  by  4  men 
placed  side  by  side  ? 

28.  How  many  different  arrangements  may  be  made  of  11 

cricketers,  supposing  the  same  2  always  to  bowl  ? 

29.  Five  flags  of  different  colors  can  be  hoisted  either  singly, 

or  any  number  at  a  time  one  above  another.  How 
many  different  signals  can  be  made  with  them  ? 

30.  How  many  signals  can  be  made  with  5  lights  of  differ- 

ent colors,  which  can  be  displayed  either  singly,  or 
any  number  at  a  time  side  by  side,  or  one  above 
another  ? 

31.  The  number  of  permutations  of  n  things,  3  at  a  time,  is 

6  times  the  number  of  combinations,  4  at  a  time. 
Find  n. 


CHAPTER  XXIX. 

CHANCE. 

393.  Definitions.  If  an  event  can  happen  in  a  ways  and 
fail  in  h  ways,  and  all  these  a-\-h  ways  are  equally  likely 
to  occur ;  if,  also,  one,  and  only  one,  of  these  a-\-h  ways 
can  occur,  and  one  'mu8t  occur;   then,  the  chance  of  the 

event  happening  is ,  and  the  chance  of  the  event  fail- 

■       .       h  rr 

mq  IS -•     rience, 

I.  The  chance  of  an  event  happening  is  expressed  by  the 
fraction  of  which  the  numerator  is  the  number  of  favorable 
ways,  and  the  denominator  the  whole  number  of  ways  favor- 
able and  unfavorable. 

Thus,  if  1  ball  is  drawn  from  a  bag  containing  3  white  balls  and 
9  black  balls,  the  chance  of  drawing  a  white  ball  is  ^^  ;  o^*.  ^^  i^  is 
expressed,  one  chance  in  four. 

II.  The  chance  of  an  event  not  happening  is  expressed  by 
the  fraction  of  which  the  numerator  is  the  number  of  unfav- 
orable ways,  and  the  denominator  the  whole  iiumber  of  ways 
favorable  and  unfavorable. 

Thus,  if  1  ball  is  drawn  from  a  bag  containing  3  white  and  9  black 
balls,  the  chance  that  it  is  not  a  white  ball  is  ■^^. 

Again,  since  — 1-  — ^—  =  1, 

a  -^h      a  +  h 

we  have  =  1  — 


a  +  6  a  +  6 

•Hence,  \i p  is  the  chance  of  an  event  happening,  \—p 
is  the  chance  of  the  event  failing. 


378  ALGEBRA. 

394.  Certainty.  If  the  event  is  certain  to  happen,  there 
are  no  ways  of  failing,  and  1— p  =  0,  therefore  ^  =  1. 
Hence  certainty  is  expressed  by  1. 

395.  Odds.  If  a  denote  the  favorable  and  b  the  unfavor- 
able ways  of  an  event,  the  odds  are  said  to  be  a  to  5  in 
favor  of  the  event,  if  -a  is  greater  than  h  ;  and  bio  a  against 
the  event,  if  b  is  greater  than  a ;  and  even  on  the  event,  if 
a  is  equal  to  b. 

396.  Examples.     Simple  Events. 

(1)  What  is  the  chance  of  throwing  double  sixes  in  one 
throw  with  two  dice  ? 

Each  die  may  fall  in  6  ways,  and  all  these  ways  are  equally 
likely  to  occur.  Hence,  the  two  dice  may  fall  in  6  X  6,  or  36,  ways 
(^  376),  and  these  36  ways  are  all  equally  likely  to  occur.  More- 
over, only  one  of  the  36  ways  can  occur,  and  one  must  occur. 

There  is  only  one  way  which  will  give  double  sixes.  Hence  the 
chance  of  throwing  double  si:^es  is  ■^^. 

(2)  What  is  the  chance  of  throwing  one,  and  only  one, 
five  in  one  throw  with  two  dice  ? 

The  whole  number  of  ways,  all  equally  likely  to  occur,  in  which 
the  dice  can  fall  is  36.  In  5  of  these  36  ways  the  first  die  will  be  a 
five,  and  the  second  die  not  a  five ;  in  5  of  these  36  ways  the  second 
die  will  be  a  five,  and  the  first  not  a  five.  Hence,  in  10  of  these  ways 
one  die,  and  only  one  die,  will  be  a  five ;  and  the  required  chance  is 


A. 


Hence,  the  odds  are  13  to  5  against  the  event. 


(3)  In  the  same  problem,  what  is  the  chance  of  throw- 
ing at  least  one  five  ? 

Here  we  have  to  include  also  the  way  in  which  both  dice  fall 
fives,  and  the  required  chance  is  \\. 

(4)  What  is  the  chance  of  throwing  a  total  of  5  in  one 
throw  with  two  dice  ? 

The  whole  number  of  ways,  all  equally  likely  to  occur,  in  which 
the  dice  can  fall  is  36.  Of  these  ways  4  give  a  total  of  5 ;  viz.,  1  and 
4,  2  and  3,  3  and  2,  4  and  1.     Hence,  the  required  chance  is  /^,  or  |. 


CHANCE.  379 

(5)  From  an  urn  containing  5  black  and  4  white  balls, 
3  balls  are  to  be  drawn  at  random.     Find  the  chance  that 

2  balls  will  be  black  and  1  white. 

There  are  9  balls  in  the  urn.    The  whole  number  of  ways  in  which 

3  balls  can  be  selected  from  9  is  ?->11A7  ^r  84. 

1x2x3 

5x4 
From  the  5  black  balls  2  can  be  selected  in  ,  or  10  ways : 

1X2'  ^    ' 

from  the  4  white  balls  1  can  be  selected  in  4  ways ;  hence,  2  black 
balls  and  1  white  ball  can  be  selected  in  10  X  4,  or  40  ways. 
The  required  chance  is  ||  =  ^f . 
Therefore  the  odds  are  11  to  10  against  the  event. 

(6)  From  a  bag  containing  10  balls,  4  are  drawn  and 
replaced ;  then  6  are  drawn.  Find  the  chance  that  the  4 
first  drawn  are  among  the  6  last  drawn. 

The  second  drawing  could  be  made  altogether  in 

110 
■!==^  =  210  ways. 

But  the  drawing  can  be  made  so  as  to  include  the  4  first  drawn  in 

16 
-=^  =  15  ways, 
12[4  ^' 

since  the  only  choice  consists  in  selecting  2  balls  from  the  6  not  pre- 
viously drawn.     Hence,  the  required  chance  is  i^^^  =  ^^. 

(7)  If  4  coppers  are  tossed,  what  is  the  chance  that 
exactly  2  will  turn  up  heads  ? 

Since  each  coin  may  fall  in  2  ways,  the  4  coins  may  fall  in  2*  =  16 
ways  (g  388).     The  2  coins  to  turn  up  heads  can  be  selected  from  the 

4x3 

4  coins  in  =  6  ways.   -Hence,  the  required  chance  is  y%  =  |. 

Therefore  the  odds  are  5  to  3  against  the  event. 

(8)  In  one  throw  with  two  dice  which  sum  is  more  likely 
to  be  thrown,  9  or  12  ? 

Out  of  the  36  possible  ways  of  falling,  four  give  the  sum  9  (namely, 
6  +  3,  3  -f  6,  5  +  4,  4  +  5),  and  only  one  way  gives  12  (namely,  6  +  6). 
Hence,  the  chance  of  throwing  9  la  four  times  that  of  throwing  12. 


380  ALGEBRA. 

Exercise  122. 

1.  The  chance  of  an  event  happening  is  f.     What  are 

the  odds  in  favor  of  the  event  ? 

2.  If  the  odds  are  10  to  1  against  an  event,  what  is  the 

chance  of  its  happening  ? 

3.  In  one  throw  with  a  pair  of  dice  what  number  is  most 

likely  to  be  thrown  ?     Find  the  odds  against  throw- 
ing that  number. 

4.  Find  the  chance  of  throwing  doublets  in  one  throw 

with  a  pair  of  dice. 

5.  If  10  persons  stand  in  a  line,  what  is  the  chance  that 

2  assigned  persons  will  stand  together  ? 

6.  If  10  persons  form  a  ring,  what  is  the  chance  that  2 

assigned  persons  will  stand  together  ? 

7.  Three  balls  are  to  be  drawn  from  an  urn  containing 

5  black,  3  red,  and  2  white  balls.     What  is  the 
chance  of  drawing  1  red  and  2  black  balls  ? 

8.  In  a  bag  are  5  white  and  4  black  balls.     If  4  balls 

are  drawn  out,  what  is  the  chance  that  they  will  be 
all  of  the  same  color  ? 

9.  If  2  tickets  are  drawn  from  a  package  of  20  tickets 

marked  1,  2,  3, ,  what  is  the  chance  that  both 

will  be  marked  with  odd  numbers  ? 

10.  From  a  bag  containing  3  white,  4  black,  and  5  red 

balls,  3  balls  are  drawn.    'Find  the  odds  against 
the  3  being  of  three  different  colors. 

11.  There  are  10  tickets  numbered  1,  2, 9,  0.     Three 

tickets  are  drawn  at  random.     Find  the  chance  of 
■  drawing  a  total  of  22. 

Note.     For  a  more  extended  discussion  of  probability,  see  Went- 
worth's  College  Algebra 


CHAPTER  XXX. 
CONTINUED   FRACTIONS. 
397.   A  fraction  in  the  form 


^/+  etc. 
is  called  a  continued  fraction,  though  the  term  is  comjuonly 
restricted  to  a  continued  fraction  that  has  1  for  each  of  its 
numerators,  as 

1 


r  +  etc. 
"We  shall  consider  in  this  chapter  some  of  the  elemen- 
tary properties  of  such  fractions. 

398.    Any  proper  fraction  in  its  lowest  terms  may  he  con- 
verted into  a  terminated  continued  fraction. 

Let  -  be  such  a  fraction ;  then,  if  p  is  the  quotient  and 
a 

c  the  remainder  of  a-^b, 

b_l_     1 
a     a  ,  c ' 

b  p+b 

if  q  is  the  quotient  and  d  the  remainder  oi  b-^c, 
1     _     1     _       1 
.  c          ,1  ,1 

P+T      P+T      p-h 


b      "^    '    b      ^    '         ,   d 


382  ALGEBRA. 

Hence,  d  =  —^ 

r  +  etc. 

The  successive  steps  of  the  process  are  the  same  as  the 
steps  for  finding  the  H.  C.  F.  of  a  and  b ;  and  since  a  and 
b  are  prime  to  each  other,  a  remainder,  1,  will  at  length 
be  reached,  and  the  fraction  terminates. 

Observe  that^,  q,  r, ,  are  all  positive  integers. 

399.  Oonvergents.  The  fractions  formed  by  taking  one, 
two,  three, ,  of  the  quotients^,  q,  r, ,  are 

111 

p  ~Ti'  ~1' 

which  simplified  are 

1         q  yr  +  1 

p    pq-\-'\^    {pq-\-l)r-\-p 

and  are  called  the  first,  second,  and  third  convergents, 

respectively. 

400.  The  successive  convergents  are  alternately  greater 
than  and  less  than  the  true  value  of  the  given  fraction. 

Let  X  be  the  true  value  of 
1 


P 


1 

^"'"r  +  etc; 


then,  since  ^,  q^  r, ,  are  positive  integers, 

1 


P<P^- 


1 
^~^r-f-etc. 


CONTINUED   FRACTIONS.  383 


-  > i— - ;  that  is,  ->x. 


^"^r  +  etc. 
Again,  ^<^  +  -4_etc. 


^^r  +  etc. 
1.1 


;>  +  -    p-\-    ^' 


^^r  +  etc; 
that  is, <  X ;  and  so  on. 

401.  If  — ,  — ,  —  are  any  three  consecutive  convergents, 

Vl       V2       V3 

and  if  mi,  m^,  ma  be  the  quotients  that  produced  them,  then 

Va'-msVj+Vi* 

For,  if  the  first  three  quotients  are  p,  q,  r,  the  first  three 
convergents  are  (§  399), 

1,        g  g^  +  1        (1) 

y  i?g+i'  (i?2'+i>'+i> 

From  (§  399)  it  is  seen  that  the  sexiond  convergent  is 
formed  from  the  first  by  writing  in  it  jo  +  -  for  j9  ;  and  the 

third  from  the  second  by  writing  q-{--  for  q.  In  this  way, 
any  convergent  may  be  formed  from  the  preceding  con- 
vergent. 


384  ALGEBRA. 

Therefore,  —  will  be  formed  from  —  by  writing  m,  H 

V3  V2  W3 

for  m^. 

In  (1)  it  is  seen  that  the  third  convergent  has  its  numer- 
ator =  r  X  (second  numerator)  +  (first  numerator)  ;  and  its 
denominator  =  r  X  (second  denominator)  +  (first  denomi- 
nator). 

Assume  that  this  law  holds  true  for  the  third  of  the 
three  consecutive  convergents 

!i«,    !^,    !i^    so  that,    !fH^^^^i  +  ^°.  (2) 

Vo     Vi     V2  V2     m^Vi  +  Vo 

Then,  since  —  is  formed  from  —  by  using  m2-\ for  mj, 


(  m2  -I ]u^ 


°  _  0713  (m-jUi  4-  ^o)  +  Uy 


Substitute  u^  and  v^  for  their  values  WaWi-fwo  and 
maVi  +  ^ol  then 

Wg W3W2  +  Wi 

Therefore  the  law  still  holds  true ;  and  as  it  has  been 
shown  to  be  true  for  the  third  convergent,  the  law  is  gen- 
eral. 


402.    The  difference  between  two  consecutive  convergents, 

Ui       7  Ho  •     1 

—  and  —  ^s 

Vi  Vj        V1V2 

-  The  difference  between  the  first  two  convergents  is 
1  Q  1 


j>     pq  +  l     pipq  +  l) 


CONTINUED    FRACTIONS.  .385 

Let  the  sign  ~  mean  the  difference  between,  and  assume 


Wo 
Wo         ^1         UqVi'^UiVo  1 


the  proposition  true  for  —  and  — 


then 


Vo        Vi  VqVi  VqVi 

But 

Wj  Wl  _  '^^2^1  '^  ^1^2  _  (^2^1  +  Wo)  i^i  '^  Wi  (^2^;]  +  ^o) 

v-i      t'l  t;i?;2  -yi^i 

(substituting  for   u^   and   Vj   their  values,  m^Ui-\-UQ  and 


Reducing, 


W2     Wi  _  Wo^?l  '^  UiVq 

V2        Vi  ViV.i 


=  —  (by  assumption). 

Hence,  if  the  proposition  be  true  for  one  pair  of  consecu- 
tive convergents,  it  will  be  true  for  the  next  pair ;  but  it 
has  been  shown  to  be  true  for  the  first  pair,  therefore  it  is 
true  for  every  pair. 

Since  by  §  400  the  true  value  of  x  lies  between  two  con- 
secutive  convergents,  —   and  — ,  the   convergent  —  will 

Vi  v^  Vi 

differ  from  a;  by  a  number  less  than  !ii  ^  ^f? .  ^-^^^^  ^g^  -^j  g^ 

Vi        V2 

number  less  than  —  ;  so  that  the  error  in  taking  —  for  x 

V1V2  Vi 

is  less  than  — ,  and  therefore  less  than  — .  as  V2  >  Vi  since 

V2  =  m^Vi  +  Vq. 

Any  convergent,  — ,  is  in  its  lowest  terms ;  for,  if  Ui  and 

Vi  had  any  common  factor,  it  would  also  be  a  factor  of 
UiVi'^UiV^ ;  that  is,  a  factor  of  1. 


386  ALGEBRA. 

403,  The  successive  convergents  approach  more  cmd  more 
nearly  to  the  true  value  of  the  continued  fraction. 

Let  — ,    — ,    —  be  consecutive  convergents. 

Vq      v^      V2 

Now  —  differs  from  x,  the  true  value  of  the  fraction, 

Vi 

only  because  Wa  is  used  instead  of  m^  -j ; — - — 

W3  +  etc. 

Let  this  complete  quotient,  which  is  always  greater  than 

unity,  be  represented  by  M. 

Vi         TUiVi  +  Vo  MVi  +  Vq 

Ux        MUl  +  Wo        U]         UqVi  ~  UiVq  1 

Vi      Mvi  +  Vo      Vi      Vi(Mvi+Vo),     'Vi(Mvi-\-Vo) 
And 

Uo        Uo     Mux  -j-  Uq M{uQVi'^UyV^ M 

Vo  ^'o       Mvx  +  ^^0        Vq{Mvx  4-  ^o)       Vq{MVx  +  v^ 

Now  \<M  and  v^  >  Vq,  and  for  both  these  reasons 

Ux  ^'?^o 
Vx        Vo 

That  is,  —  is  nearer  to  x  than  —  is. 

Vi  Vo 


404.  Any  convergent  —  is  nearer  the  true  value  x  than 
any  other  fraction  with  smaller  denominator. 

Let  -  be  a  fraction  in  which  b<,Vx. 
o 

If  ^  is  one  of  the  convergents,  x^~>'^^x.         §  403 

0  0        Vx 

If  -  is  not  one  of  the  convergents,  and  is  nearer  to  x 


CONTINUED   FRACTIONS.  387 


than  —  is,  then,  since  x  lies  between  —  and  —  (§  400), 
Y  must  be  nearer  to  —  than  —  is ;  that  is, 

0  V2  Vi 

a      U2  ^u,      U2       v^a  '^  uJb        1 
o      Vi      Vi      V2  V2O  yii;2 

and  since  h  <  v^,  this  would  require  that   v^a  '^  Uih  <  1 ; 

but  v^cn^uj}  cannot  be  less  than  1,  for  a,  h,  2/2,  '^2  are  all 

integers.     Hence,  —  is  nearer  to  x  than  7  is. 
Vi  h 

405.  Find  the  continued  fraction  equal  to  fj-,  and  also 
the  successive  convergents. 

Following  the  process  of  finding  the  H.  C.  F.  of  31  and  75,  the 
Buccessive  quotients  are  found  to  be  2,  2,  2,  1,  1,  2.  Hence  the  con- 
tinued fraction  is 

1 


To  find  the  successive  convergents : 

Write  the  successive  quotients  in  line,  \  under  the  first  quotient, 
\  under  the  second  quotient,  and  then  (§  401)  multiply  each  term  by 
the  quotient  above  it,  and  add  the  term  to  the  left  to  obtain  the 
corresponding  term  to  the  right.     Thus, 

Quotients      =  2,  2,  2,   1,    1,    2. 
Convergents  =  |,  |,  f ,  x\,  ^^,  \%  f f 
It  is  convenient  to  begin  to  reckon  with  £,  but  the  next  conver- 
gent, in  this  case  \,  is  called  the /rsi  convergent. 

Note.  Continued  fractions  are  often  written  in  a  more  compact 
form.    Thus,  the  above  fraction  may  be  written 

1     1     1     1     1     1 

2+2+2+1+1+2 


388  ALGEBRA. 


406.  A  quadratic  surd  may  be  expressed  in  the  form  of 
a  non-terminating  continued  fraction. 

To  express  V3  in  the  form  of  a  continued  fraction. 

Vs  =  1  +  -  (as  1  is  the  greatest  integer  in  Vs) ; 


then  1_V3-1. 

X 

1         Vs  +  l 


V3-1  2 


2 


=  1  +  -  (  as  1  is  the  greatest  integer  in 

y\ 


then  l=VLti_i  =  V3^: 

y  2  2 

_.  2  V3  +  1 


V3-1  1 


Suppose ^  =  2  +  -  (  as  2  is  the  greatest  integer  in  — J 


then  l:.Vl±I_2=V3-l. 

2  1 


V3-1 

This  is  the  same  as  x  above ;  hence,  the  quotients  1,  2,  will  be 
continually  repeated. 

1 


^  ^  2  +  etc. 


.-.  V3  =  1  + 

of  which will  be  continually  repeated,  and  the  whole  expres- 

Kion  may  be  written, 

^1+2 
The  convergentB  will  be  1,  2,  f ,  |,  }f,  ff ,  |j,  etc. 


CONTINUED    FRACTIONS.  389 

407.  A  continued  fraction  in  which  the  denominators 
recur  is  called  a  periodic  continued  fraction. 

The  value  of  a  periodic  continued  fraction  can  be  ex- 
pressed as  the  root  of  a  quadratic  equation. 

Find  the  surd  value  of 


A.    XJJIV 

1-1- 

-2 

Let  X  be  the 

value ; 

fVipn 

x^ 

1 

_2  +  aj. 

tXicU 

-^ 

3+a;' 

.-.  x'  +  2x 

=  2, 

X 

=  -  1  +  V3. 

We  take  the 

+  sign 

since  x  is  evidently  positive. 

408.    An  exponential  equation  can  be  solved   by  con- 
tinued fractions. 

Solve  by  continued  fractions  10*  —  2. 
1. 


1 

then  IQy  =  2, 

or  10  =  2*. 


.-.  y  =  3  +  i  (as  10  lies  between  2^  and  2*). 

3+1  I 

Then  10  =  2    "  =  23x2*; 

1 
or  2^=-V-  =  |. 

and  2  =  (I)-. 

...  z  =  3  +  -  fas  2  lies  between  /"-Y  and  /'-Yl. 

8+1  i 

Then  2  =  (fr«  =  (|)»X(f)«; 

or  (i)"  =  m 

and  f-(mr. 


390  ALGEBRA. 

The  greatest  integer  in  u  will  be  found  to  be  9. 

1 


Hence,  a;  =  0  + 


3+^ 


1 

o  i — 

9  +  etc. 
The  successive  convergents  will  be  ^,  ^jj,  f f ,  etc. 
The  last  gives         a?  =  f  f  =  0.3010,  nearly. 

Observe  that  by  the  above  process  we  have  calculated  the  common 
logarithm  of  2.     By  ^  402  the  error,  when  0.3010  is  -taken  for  the 

common  logarithm  of  2,  is  considerably  less  than ;  that  is,  con- 
siderably less  than  0.00011 ;  so  that  0.3010  is  certainly  correct  to 
three  places  of  decimals,  and  probably  correct  to  four  places. 

Logarithms  are,  however,  much   more   easily  calculated  by  the 
use  of  series,  as  will  be  shown  in  a  following  chapter. 


Exercise  123. 

1.  Find  continued   fractions  for   -J^-f-;   ^j-;   V5;   Vll ; 

4V6 ;  and  find  the  fourth  convergent  to  each. 

2.  Find  continued  fractions  for  ^;   HI;   ffff;  ^J-; 

and  find  the  third  convergent  to  each. 

3.  Find  continued  fractions  for  V2l ;   V22;   V33;  V55. 

4.  Obtain  convergents,  with  only  two  figures  in  the  denom- 

inator, that  approach  nearest  to  the  values  of  VlO ; 
Vl5;   VTZ;   Vl8 ;  V20. 

5.  Find   the   proper  fraction  which,  if  converted  into  a 

continued  fraction,  will  have  quotients  1,  7,  5,  2. 

6.  Find  the  next  convergent  when  the  two  preceding  con- 

vergents are  j^  and  -1^,  and  the  next  quotient  is  5. 

7.  Find  a  series  of  fractions  to  the  ratio  of  a  yard  to  a 

meter,  if  a  meter  equal  1.0936  yards. 


CONTINUED    FRACTIONS.  391 

8.  If  the  pound  troy  is  the  weight  of  22.8157  inches  of 

water,  and  the  pound  avoirdupois  of  27.7274  inches, 
find  a  fraction  with  denominator  <  100  which  shall 
difi'er  from  their  ratio  by  <  0.0001. 

9.  The  ratio  of  the  diagonal  to  a  side  of  a  square  being 

V2,  find  a  fraction  with  denominator  <  100  which 
shall  differ  from  their  ratio  by  <  0.0001. 

10.  The  ratio  of  the  circumference  of  a  circle  to  its  diame- 

ter being  3.14159265,  find  the  first  three  convergents, 
and  determine  to  how  many  decimal  places  each  may 
be  depended  upon  as  agreeing  with  the  true  value. 

11.  Two  scales  whose  zero  points  coincide  have  the  dis- 

tances between  consecutive  divisions  of  the  one  to 
those  of  the  other  as  1  :  1.06577.  Find  what  divis- 
ion-points most  nearly  coincide. 

12.  Find  the  surd  values  of 

o.l    i     i    1    i     .  ,i    1    i 

"^1  +  6'     3  +  1  +  6'        "*'2  +  3  +  4 

13.  Show  that  the  ratio  of  the  diagonal  of  a  cube  to  its 

edge  may  be  nearly  expressed  by  97  :  56.  Find  the 
limit  of  the  error  made  in  taking  this  ratio  for  the 
true  ratio. 

14.  Find  a  series  of  fractions  converging  to  the  ratio  of  5 

hours  48  minutes  51  seconds  to  24  hours. 

15.  Find  a  series  of  fractions  converging  to  the  ratio  of  a 

cubic  yard  to  a  cubic  meter,  if  1  cubic  yard  =  /oVoVo 
of  a  cubic  meter. 


CHAPTER   XXXI. 
SCALES   OF   NOTATION. 

409.  Definitions.  Let  any  positive  integer  be  selected  as 
a  radix  or  base ;  then  any  number  may  be  expressed  as  a 
polynomial  of  which  the  terms  are  multiples  of  powers  of 
the  radix. 

Any  positive  integer  may  be  selected  as  the  radix ;  and 
to  each  radix  corresponds  a  scale  of  notation. 

In  writing  the  polynomials  they  are  arranged  by  descend- 
ing powers  of  the  radix,  and  the  powers  of  the  radix  are 
omitted,  the  place  of  each  digit  indicating  of  what  power 
of  the  radix  it  is  the  coefficient. 

Thus,  in  the  scale  of  ten,  2356  stands  for 

2xl0'  +  3x  102 +  5x10  +  6; 
in  the  scale  of  seven  for 

2x7^  +  3x72  +  5x7  +  6; 
in  the  scale  of  r  for 

2r3  +  3r2  +  5r  +  6. 

410.  Computation.  Computations  are  made  with  numbers 
in  any  scale,  by  observing  that  one  unit  of  any  order  is 
equal  to  the  radix-number  of  units  of  the  next  lower  order ; 
and  that  the  radix-number  of  units  of  any  order  is  equal 
to  one  unit  of  the  next  higher  order. 

(1)  Add  56,432  and  15,646  (scale  of  seven). 

56432  '^^^  process  differs  from  that  in  the  decimal  scale 

15646  ®°V  ^^  that  when  a  sum  greater  than  seven  is  reached, 

we  divide  by  seven  (not  ten),  set  down  the  remainder, 

105411  ^^^  carry  the  quotient  to  the  next  column. 


SCALES    OF   NOTATION.  393 

(2)  Subtract  34,561  from  61,235  (scale  of  eight). 

61235 

34561  ^®  ^^^  ^ight,  instead  of  ten  as  in  the  common 

24454       ^^^^®- 

(3)  Multiply  5732  by  428  (scale  of  nine). 

5732 
428 


51477        We  divide  each  time  by  nine,  set  down  the  remain- 
12564      der,  and  carry  the  quotient. 
24238 

2612127 

(4)  Divide  2,612,127  by  5732  (scale  of  nine). 

5732)2612127(428 
24238 


17732 
12564 

51477 
51477 


411.  Integers  in  Any  Scale.  If  i  be  any  positive  integer, 
any  positive  integer  N"  may  he  expressed  in  the  form 

JSr=  ar""  +  W-^  + ■\-pr''  4-^^  +  5, 

in  which  the  coefficients  a,  b,  c, ,  are  positive  integers,  each 

less  than  r. 

For,  divide  iVby  r"*,  the  highest  power  of  r  contained  in 
iV,  and  let  the  quotient  be  a  with  the  remainder  iVj. 

Then,  N=ar'^-\-N^. 

In  like  manner,  iVi  =  Sr^-^  +  ^2 ;  iV,  =  c?-""'' +  iVj ; 
and  so  on. 

By  continuing  this  process,  a  remainder  s  will  at  length 
be  reached  which  is  less  than  r.     So  that, 

N^  ar""  4-  br""^  -\- +^r^  +  jr  +  s. 


394  ALGEBRA. 

Some  of  the  coefficients  s,  q,  p, may  vanish,  and  each 

will  be  less  than  r ;  that  is,  their  values  may  range  from 
zero  to  r — 1.  Hence,  including  zero,  r  digits  will  be 
required  to  express  numbers  in  the  scale  of  r. 

To  express  any  positive  integer  N  in  the  scale  of  r. 
~    It  is  required  to  express  N  in  the  form 

ar**  +  5r"~^  + +^r^  -\- qr -\- s, 

and  to  show  how  the  digits  a,  5, may  be  found. 

If  N=  ar""  +  Z>r~-^  + -\- pr""  -f-  qr  +  5, 

JSF  s 

then  —  =  ar"-^  +  hr*"-'  + +^r  +  g  +  - 

That  is,  the  remainder  on  dividing  iV  by  r  is  s,  the  last 
digit. 

Let  iVi  =  ar""-'  +  br""-'  + +pr  +  q ; 

then  —  =  ar""-'  +  br""'^  + +  P  +  -• 

That  is,  the  remainder  is  q,  the  last  but  one  of  the  digits. 

Hence,  to  express  an  integral  number  in  a  proposed  scale, 

Divide  the  number  by  the  radix,  then  the  quotient  by  the 

radix,  and  so  on ;  the  successive  remainders  will  be  the 

successive  digits  beginning  with  the  units'  place. 

(1)  Express  42,897  (scale  of  ten)  in  the  scale  of  six. 

6)42897 
6)7149. ...  3 

6)1191 3 

6)198 3 

6)33.  ...  0 

5 3 

-4w5.  530,333.    * 


SCALES   OF   NOTATION. 


395 


(2)  Change  37,214  from  the  scale  of  eight  to  the  scale  of 
nine. 

The  radix  is  8 ;  and  hence  the  two  digits  on  the 
9)37214  left,  37,  do  not  mean  thirty-seven,  but  3x8  +  7, 

9}_3363  ...  1     or  thirty -one,  which  contains  9  three  times,  with 
9]305.  ...  6     t,he  remainder  4. 

9)25 8        The   next  partial   dividend   is  4  x  8  +  2  =  34, 

2.  ...  3     which  contains  9  three  times,  with  the  remainder 
Ans.  23,861.     7.  and  so  on. 

(3)  In  what  scale  is  140  (scale  of  ten)  expressed  by  352  ? 
Let  r  be  the  radix ;  then,  in  the  scale  of  ten, 

140  =  3r2  +  5r  +  2    or    3r2  +  5r=138. 

Solving,  we  find  r  =  6. 

The  other  value  of  r  is  fractional,  and  therefore  inadmissible,  since 
the  radix  is  always  a  positive  integer. 

412.  Kadix-Practions.     As  in  the  decimal  scale  decimal 
fractions  are  used,  so  in  any  scale  radix-fractions  are  used. 

Thus,  in  the  decimal  scale,  0.2341  stands  for 

10       102        103       104' 

and  in  the  scale  of  r  it  stands  for 

2  +  ^+i  +  l. 
r^r^^r^      r*- 

(1)    Express  ||^|  (scale  of  ten)  by  a  radix-fraction  in  the 
scale  of  eight. 

Assume  245^a     _&       c,      rf      

256     8      82     83     8* 

Multiply  by  8.         7f  ^  =  a  +  ^  + 1  +  |  + 

...  a  =  7.  and  21^&      c_^     

32     8     8"     &» 


396 


6 

ALGEBRA. 

Multiply  by  8, 

^i-^+M- 

.-.  &=-5,  and 

i  =  £  +  i  + 

4     8     8' 

Multiply  by  8, 

^  =  —8  + 

.-.  c  =  2,  and' 

0  =  d,  etc. 

The  answer  is  i 

0.752. 

We  take  the  ir 
separately. 

itegral  and  fractional 

parts 

Integral  part  : 

6)35 
4 

5. 

Fractional  part . 

•    h 

4  _12 

82     64 

3 

16' 

(2)  Change  35.14  from  the  scale  of  eight  to  the  scale  of 

six. 

3 

16)18(1 
16 
2 
_6 

16)12(0 

_6 
16)72(4 
64 
8 

This  is  reduced  to  a  radix-fraction  in  the  scale  6 

of  six  as  follows  :  1 6)  48  (3 

48 

45.1043 
Exercise  124. 

1.  If  6,  7,  8,  3,  2  are  the  digits  of  a  number  in  the  scale  of 

r,  beginning  from  the  right,  write  the  number. 

2.  Find  the  product  of  234  and  125  when  r  is  the  base. 

3.  In  what  scale  will  756  be  expressed  by  530? 

4.  In  what  scale  will  540  be  the  square  of  23  ? 

5.  In  what  scale  will  212,  1101,  1220  be  in  arithmetical 

progression  ? 

6.  Multiply  31.24  by  0.31  in  the  scale  of  5. 


CHAPTER  XXXII. 
THEORY   OF   NUMBERS. 

413.  Definitions.  In  the  present  chapter,  by  number  will 
be  me&ut  positive  integer.  The  terma  prime,  composite,  will 
be  used  in  the  ordinary  arithmetical  sense. 

A  multiple  of  a  is  a  number  which  contains  the  factor  a, 
and  may  be  written  ma. 

An  even  number,  since  it  contains  the  factor  2,  may  be 
written  2m;  an  odd  number  may  be  written  2m +  1, 
2m-l,  2m-}-3,  2m-3,  etc. 

A  number  a  is  said  to  divide  another  number  b  when 

b  .         .  , 

-  is  an  integer. 

a 

414.  Eesolution  into  Prime  Pactors.  A  number  can  be 
resolved  into  prime  factors  in  only  one  way. 

Let  iV  be  the  number;    suppose    JSf—abc ,   where 

a,b,c, are  prime  numbers;    suppose  also  iV=a)Sy 

where  a,  p,  y, are  prime  numbers. 

Then,  abc =al3y 

Hence,  a  must  divide  the  product  abc ;  but  a,  b,  c, 

are  all  prime  numbers  ;  hence  a  must  be  equal  to  some  one 
of  them,  a  suppose. 

Dividing  by  a,  be =  Py , 

and  so  on.    Hence,  the  factors  in  a/3y are  equal  to  those 

in  abc ,  and  the  theorem  is  proved. 

415.  Divisibility  of  a  Product.  1.  If  a  number  a  divides  a 
product  bo,  and  is  prime  to  b|  it  mu^t  divide  c 


398  ALGEBRA. 

For,  since  a  divides  he,  every  prime  factor  of  a  must  be 
found  in  he ;  but  since  a  is  prime  to  h,  no  factor  of  a  will 
be  found  in  h ;  hence  all  the  prime  factors  of  a  are  found 
in  c  ;  that  is,  a  divides  c. 

From  this  theorem  it  follows  that : 

II.  If  a  prime  numher  a  divides  a  product  bode....,  it 
must  divide  somefaetor  of  that  produet ;  and  conversely. 

III.  If  a  prime  numher  divides  b",  it  must  divide  b. 

IV.  If  a  is  prime  to  b  and  c,  it  is  prime  to  be. 

Y.  If  a,  is  prime  to  b,  every  power  of  a  is  prime  to  every 
power  ofh, 

416,  Theorem.    If  - ,  a  fraction  in  its  lowest  terms,  is  equal 

to  another  fraction  -,  then  c  and  d  are  equi7nuliiples  of  a 

andh. 

If  ?  =  -,  then  ^=c.      Since  h  will  not  divide  a,  it 
b      d  b 

must  divide  d]  hence  c?  is  a  multiple  of  h. 

Let   d  =  mb,    m   being    an    integer ;    since  y  =  ~^>   ^^^ 

0      a 

d  =  mh,  -  =  — -  ;  therefore  c  =  ma. 
b      mb 

Hence,  c  and  d  are  equimultiples  of  a  and  h. 

From  the  above  theorem,  it  follows  that  in  the  decimal 
scale  of  notation  a  common  fraction  in  its  lowest  terms  will 
produce  a  non-terminating  decimal  if  its  denominator  con- 
tains any  prime  factor  except  2  and  5. 

For  a  terminating  decimal  is  equivalent  to  a  fraction  with 

a  denominator  10".     Therefore,  a  fraction  j  in  its  lowest 

6 

terms  cannot  be  equal  to  such  a  fraction,  unless  10"  is  a 
multiple  of  b.     But  10",  that  is,  2"  X  5",  contains  no  factors 


THEORY    OF   NUMBERS.  399 

besides  2  and  5,  and  hence  cannot  be  a  multiple  of  b,  if  b 
contains  any  factors  except  these. 

417.  Square  Numbers.  If  a  square  number  is  resolved  into 
its  prime  factors,  the  exponent  of  each  factor  will  be  even. 

For,  if  N  =  a^  Xh^  Xc'  , 

N^=a^pxb-"^xc''' 

Conversely :  A  number  which  has  the  exponents  of  all 
its  prime  factors  even  will  be  a  perfect  square  ;  therefore, 
to  change  any  number  to  a  perfect  square, 

Resolve  the  number  into  its  prime  factors,  select  the  fac- 
tors which  have  odd  exponents,  and  multiply  the  given 
number  by  the  product  of  these  factors. 

Thus,  to  find  the  least  number  by  which  250  must  be  multiplied  to 
make  it  a  perfect  square. 

250  =  2x5',  in  which  2  and  5  are  the  factors  which  have  odd 
exponents. 

Hence  the  multiplier  required  is  2  X  5=  10. 

418.  Divisibility  of  Numbers. 

I.  If  two  numbers  N  and  N',  when  divided  by  a,  have 
the  same  remainder,  their  difference  is  divisible  by  a. 

For,  if  iVwhen  divided  by  a  have  a  quotient  q  and  a 
remainder  r,  then 

]Sf=qa^r. 

And  if  N^  when  divided  by  a  have  a  quotient  q^  and  a 
remainder  r,  then 

iV'=  q^a  +  r. 

Therefore,  N-N'=-{q  —  q')  a. 

II.  If  the  difference  of  two  numbers  N  arid  W  is  divisible 
by  a,  then  N  and  W  when  divided  by  a  will  have  the  same 
remainder. 


400 

ALGEBRA. 

For,  if 

iV^- 

-JV'  =  (q-q') 

then 

]sr_ 

a 

Therefore, 

a 

JST'       , 
-q     = — -q\ 

That  is, 

jsr- 

-  aq  =  W—  ag 

III.  ^  two  numbers  N  and  N',  when  divided  by  a  given 
number  a,  have  o-emainders  r  and  r',  then  UN'  and  rr'  when 
divided  by  a  will  have  the  same  remainder. 

For,  if  N  =qa-\-r, 

and  ]Sr^q'a-\-r\ 

then  iViV'  =  qq^a^-\-  qar'-j-  q'ar+rr^ 

=  (qq'a-]-  qr'-{-  q'r)  a + r?'. 

Therefore,  by  II.,  JYJV'  and  rr'  when  divided  by  a  will 
have  the  same  remainder. 

As  a  particular  case,  37  and  47  when  divided  by  7  have  remainders 
2  and  5  respectively. 

Now  37  X  47  =  1739  and  2  x  5  =  10. 

The  remainder,  when  each  of  these  two  numbers  is  divided  by  7, 
is  3. 

Note.   From  II.  it  follows  that,  in  the  scale  of  ten : 

(1)  A  number  is  divisible  by  2,  4,  8, if  the  numbers  denoted 

by  its  last  digit,  last  two  digits,  last  three  digits are  divisible 

respectively  by  2,  4,  8, 

(2)  A  number  is  divisible  by  5,  25,  125, if  the  numbers  denoted 

by  its  last  digit,  last  two  digits,  last  three  digits, are  divisible 

respectively  by  5,  25,  125 

(3)  If  from  a  number  the  sum  of  its  digits  is  subtracted,  the 
remainder  will  be  divisible  by  9. 


THEORY    OF   NUMBERS.  401 

For,  if  from  a  number  expressed  in  the  form 

a  +  10&  +  102c  +  103(^+ 

a+      b  +       c+       d  + is  subtracted, 

the  remainder  will  be  (10  -1)6  +  (10^  - 1)  c  +  (10'  -l)d  + 

and  10  -  1,  102  -  1,  10^  -  1, will  be  a  series  of  9's. 

Therefore,  the  remainder  is  divisible  by  9. 

(4)  Hence,  a  number  N  may  be  expressed  in  the  form 

9  n  +  s  (if  s  denotes  the  sum  of  its  digits) ; 
and  i\r  will  be  divisible  by  3  if  s  is  divisible  by  3  ;  and  also  by  9  if  s 
is  divisible  by  9. 

(5)  A  number  will  be  divisible  by  11  if  the  difference  between 
the  sum  of  its  digits  in  the  even  places  and  the  sum  of  its  digits  in 
the  odd  places  is  0  or  a  multiple  of  11. 

For,  a  number  N  expressed  by  digits  (beginning  from  the  right) 
a,  6,  c,  d, may  be  put  in  the  form  of 

iV=  a  +  105  +  102c  +  10^  d  + 

...  x^-a+b-c  +  d- =  (10  +  l)5  +  (102-l)c  +  (103  +  l)c?  + 

But  10  +  1  is  a  factor  of  10  +  1,  10^  -  1,  10^  +  1, 

Therefore,  N-  a  +  b  —  c  +  d  — is  divisible  by  10  +  1  =  11.  . 

Hence,  the  number  iVmay  be  expressed  in  the  form 
11  w  +  (a  +  c  + )  -{b  +  d+ ), 

and  will  be  a  multiple  of  11  if  (a  +  c  + )  —  {b  +  d  + )  is  0  or  a 

multiple  of  11. 

419.  Theorem.  The  product  of  r  consecutive  integers  is 
divisible  5y  |r. 

Kepresent  by  P„,  „  the  product  of  h  consecutive  integers 
beginning  with  n. 

Then,        P„,  u  =  'n(n  +  1) (n  +  k-l); 

P„+,,*+,  =  (n  +  l)(n  +  2) (n  +  kXn  +  ki-l) 

=  n{n  +  V){n  +  2) (n  +  k) 

+  (^  +  l)(n  +  l)(^  +  2) {n-^k), 

.-.   Pn+l.  »+i  =  Pn. *+i  +  (^  +  1)  P«+i.*. 


402  ALGEBRA. 

Assume,  for  the  moment,  that  the  product  of  any  k  con- 
secutive integers  is  divisible  by  [k. 

Then,  P,+,.,+,  -  P,,,^,  -\-{]c-\-l)M\h- 

or  P„+,, ,+1  -  P„. ,+1  +  M\]c±l ; 

where  ilf  is  an  integer. 

From  this  it  is  seen  that  if  Pn,k+\  is  divisible  by  1^+  1. 
P„+i,jfe+i  is  also  divisible  by  |/:-|-  1 ;  but  Pi,*+i  is  divisible 
by  1^+1  since  Pi.^+i  =1^+1-  •"•  P-i.k+i  is  divisible  by 
1^  +  1 ;  .".  P-i,k^\  is  divisible  by  1^  +  1 ;  and  so  on. 

Hence,  the  product  of  any  ]c-\-\  consecutive  integers  is 
divisible  by  |^+  1,  if  it  is  known  that  the  product  of  any  k 
consecutive  integers  is  divisible  by  \k.  But  the  product  of 
any  2  consecutive  integers  is  divisible  by  [2 ;  therefore,  the 
product  of  any  3  consecutive  integers  is  divisible  by  |3^; 
therefore,  the  product  of  any  4  consecutive  integers  is 
divisible  by  |^;  and  so  on.  Therefore,  the  product  of  any 
r  consecutive  integers  is  divisible  by  [r. 


Exercise  125. 

Find  the  least  number  by  which  each  of  the  following 
numbers  must  be  multiplied  in  order  that  the  product  may 
be  a  square  number. 

1.    2625.  2.   3675.  3.   4374.  4.    74088. 

6.    If  w  and  n  are  positive  integers,  both  odd  or  both 
even,  show  that  m"^  —  n^  is  divisible  by  4. 

6.  Show  that  n^  —  ni?,  always  even. 

7.  Show  that  n^  —  w  is  divisible  by  6  if  w  is  even  ;  and 

by  24  if  w  is  odd. 


CHAPTER  XXXIII. 

VARIABLES  AND   LIMITS. 

420.  Oonstants  and  Variables.  A  number  that,  under  the 
conditions  of  the  problem  into  which  it  enters,  may  be  made 
to  assume  any  one  of  an  unlimited  number  of  values  is 
called  a  variable. 

A  number  that,  under  the  conditions  of  the  problem  into 
which  it  enters,  has  a  fixed  value  is  called  a  constant. 

Variables  are  generally  represented  by  x,  y,  z,  etc. ;  con- 
stants, by  the  Arabic  numerals,  and  by  a,  h,  c,  etc. 

421.  Functions.  Two  variables  may  be  so  related  that  a 
change  in  the  value  of  one  produces  a  change  in  the  value 
of  the  other.  In  this  case  one  variable  is  said  to  be  a 
function  of  the  other. 

Thus,  if  a  man  walks  on  a  straight  road  at  a  uniform  rate  of  a 
miles  per  hour,  the  number  of  miles  he  walks  and  the  number  of  hours 
he  walks  are  both  variables,  and  the  first  is  a  function  of  the  second. 
If  y  be  the  number  of  miles  he  has  walked  at  the  end  of  x  hours,  y 
and  X  are  connected  by  the  relation  y  =  ax,  and  3/  is  a  function  of  x. 

y 

Also  x  =  -  \  hence,  x  is  also  a  function  of  y. 

When  one  of  two  variables  is  a  function  of  the  other,  the 
relation  between  them  is  generally  expressed  by  an  equa- 
tion. If  a  value  of  one  variable  is  assumed,  the  corre- 
sponding value  of  the  other  variable  can  be  found  from  the 
given  equation  of  relation  between  the  two  variables. 

The  variable  of  which  the  value  is  assumed  is  called  the 
independent  variable  ;   the  variable  of  which  the  value  is 


404  ALGEBRA. 

found  from  the  given  relation  of  the  two  variables  is  called 
the  dependent  variable. 

In  the  last  example  we  may  assume  values  of  x,  and  find  the  cor- 
responding values  of  y  from  the  relation  y  =  ax\  or  assume  values  of 

y 

y,  and  find  the  corresponding  values  of  x  from  the  relation  x=  -.     In 

the  first  case  x  is  the  independent  variable,  and  y  the  dependent ;  in 
the  second  case  y  is  the  independent  variable,  and  x  the  dependent. 

422,  Limits.  As  a  variable  changes  its  value,  it  may- 
approach  some  constant;  if  the  variable  can  be  made  to 
approach  the  constant  as  near  as  we  please,  but  cannot  be 
made  absolutely  equal  to  the  constant,  the  variable  is  said 
to  approach  the  constant  as  a  limit,  and  the  constant  is 
called  the  limit  of  the  variable. 

Let  X  represent  the  sum  of  n  terms  of  the  infinite  series 


then(§314),  ..  (^_|^=i  =  2- J-j. 

Suppose  n  to  increase ;  then,  decreases,  and  x  approaches  2. 

Since  we  can  take  as  many  terms  of  the  series  as  we  please,  n  can 
be  made  as  large  as  we  please ;  therefore,  — —  can  be  made  as  small 

as  we  please,  and  a;  can  be  made  to  approach  2  as  near  as  we  please. 

We  cannot,  however,  make  x  absolutely  equal  to  2. 

If  we  take  any  assigned  value,  as  j-^^-q^,  we  can  make  the  dif- 
ference between   2  and  x  less  than  this  assigned  value ;   for  we 

have  only  to  take  n  so  large  that is  less  than  ^^^-q^  ;  that  is,  that 

2"-^  is  greater  than  10,000:  this  will  be  accomplished  by  taking  n 
as  large  as  15.  Similarly,  by  taking  n  large  enough,  we  can  make 
the  difference  between  2  and  x  less  than  any  assigned  value. 

Since  2  —  x  can  be  made  as  small  as  we  please,  it  follows  that  the 

sum  of  n  terms  of  the  series  lH-j  +  i  +  |  + ,  as  n  is  constantly 

increased,  approaches  2  as  a  limit. 


VARIABLES   AND   LIMITS.  405 

423.  Test  for  a  Limit.  In  order  to  prove  that  a  variable 
approaches  a  constant  as  a  limit,  it  is  necessary  and  suffi- 
cient to  prove  that  the  difference  between  the  variable  and 
the  constant  can  be  made  as  near  to  zero  as  we  please,  but 
cannot  be  made  absolutely  equal  to  zero. 

A  variable  may  approach  a  constant  without  approaching  it  as  a 
limit.  Thus,  in  the  last  example  x  approaches  3,  but  not  as  a  limit ; 
for  3  -  aj  cannot  be  made  as  near  to  0  as  we  please,  since  it  cannot 
be  made  less  than  1. 

424.  Infinites.  As  a  variable  changes  its  value,  it  may 
constantly  increase  in  numerical  value;  if  the  variable 
can  become  numerically  greater  than  any  assigned  value, 
however  great  this  assigned  value  may  be,  the  variable  is 
said  to  increase  without  limit,  or  to  increase  indefinitely. 

When  a  variable  is  conceived  to  have  a  value  greater 
than  any  assigned  value,  however  great  this  assigned  value 
may  be,  the  variable  is  said  to  become  infinite;  such  a 
variable  is  called  an  infinite  number,  or  simply  an  infinite. 

425.  Infinitesimals.  As  a  variable  changes  its  value,  it 
may  constantly  decrease  in  numerical  value  ;  if  the  vari- 
able can  become  numerically  less  than  any  assigned  value, 
however  small  this  assigned  value  may  be,  the  variable  is 
said  to  decrease  witJiout  limit,  or  to  decrease  indefinitely. 

In  this  case  the  variable  approaches  0  as  a  limit. 

When  a  variable  which  approaches  0  as  a  limit  is  con- 
ceived to  have  a  value  less  than  any  assigned  value,  how- 
ever small  this  assigned  value  may  be,  the  variable  is  said 
to  become  infinitesimal;  such  a  variable  is  called  an  infini- 
tesimal number,  or  simply  an  infinitesimal. 

426.  Infinites  and  infinitesimals  are  variables,  not  con- 
stants. There  is  no  idea  of  fixed  value  implied  in  either 
an  infinite  or  an  infinitesimal. 


406  ALGEBRA. 

An  infinitesimal  is  not  0.  An  infinitesimal  is  a  variable 
arising  from  the  division  of  a  quantity  into  a  constantly- 
increasing  number  of  parts ;  0  is  a  constant  arising  from 
taking  the  difference  of  two  equal  quantities. 

A  number  which  cannot  become  infinite  is  said  to  be 
finite. 

427.  Eelations  between  Infinites  and  Infinitesimals. 
l.Ifx  is  infinitesimal,  and  a  is  finite  and  not  0,  then  ax 

is  infinitesimal.    For,  ax  can  be  made  as  small  as  we  please 
since  x  can  be  made  as  small  as  we  please. 

II.  If  X  is  infinite,  and  a  is  finite  and  not  0,  then  aX  is 
infinite.  For  aX  can  be  made  as  large  as  we  please  since 
Xcan  be  made  as  large  as  we  please. 

III.  If  X  is  infinitesimal,  and  a  is  finite  and  not  0,  then 
-  is  infinite.  For  -  can  be  made  as  large  as  we  please 
since  x  can  be  made  as  small  as  we  please. 

IV.  If  X  is  infinite,  and  a  is  finite  and  not  0,  then  — :  is 

a 
infinitesimal.     For  "^  can  be  made  as  small  as  we  please 

since  ^can  be  made  as  large  as  we  please. 

In  the  above  theorems  a  may  be  a  constant  or  a  variable ; 

the  only  restriction  on  the  value  of  a  is  that  it  shall  not 

become  either  infinite  or  zero. 

428.  Abbreviated  Notation.  An  infinite  is  often  repre- 
sented by  00.    In  §  427,  III.  and  IV.  are  sometimes  written: 

=  00,      —  ^-=  0. 

0  on 

The  expression  -  cannot  be  interpreted  literally,  since  we  cannot 
divide  by  0 ;  and  the  expression  —  =  0  cannot  be  interpreted  liter- 


VARIABLES    AND    LIMITS.  407 

ally,  since  we  can  find  no  number  such  that  the  quotient  obtained 
by  dividing  a  by  that  number  is  zero. 

-  =  00  is  simply  an  abbreviated  way  of  writing  -.  if  -  =  X,  and  x 
0  X 

approaches  0  as  a  limit,  X  increases  without  limit.     -  =  0  is  simply 

an  abbreviated  way  of  writing :  if  —  =  x,  and  X  increases  without 

limit,  X  approaches  0  as  a  limit. 

429.  Approach  to  a  Limit.  When  a  variable  approaches  a 
limit,  it  may  approach  its  limit  in  one  of  three  ways : 

(1)  The  variable  may  be  always  less  than  its  limit. 

(2)  The  variable  may  be  always  greater  than  its  limit. 

(3)  The  variable  may  be  sometimes  less  and  sometimes 
greater  than  its  limit. 

If  a;  represents  the  sum  of  n  terms  of  the  series  l+i  +  i  +  |  + , 

X  is  always  less  than  its  limit  2. 

If  re  represents  the  sum  of  n  terms  of  the  series  3  —  |  —  |— |^  — , 

X  is  always  greater  than  its  limit  2. 

If  ic  represents  the  sum  of  n  terms  of  the  series  3  —  f  +  f  —  f  + , 

we  have  (^  314) 

1  +  1  ^     -^ 

As  n  is  indefinitely  increased,  x  evidently  approaches  2  as  a  limit. 

If  n  is  even,  x  is  less  than  2 ;  if  n  is  odd,  x  is  greater  than  2. 
Hence,  if  n  be  increased  by  taking  each  time  one  more  term,  x  will 
be  alternately  less  than  and  greater  than  2.     If,  for  example, 

w=    2,        3,        4,        5,        6,         7, 
x^.ll        2i       n,      2tV     1H.     211. 

In  whatever  way  a  variable  approaches  its  limit,  the  test 
of  §  423  always  applies. 

430.  Equal  Variables.  If  two  variables  are  equal  and  are 
so  related  that  a  change  in  the  one  produces  such  a  change 


408  ALGEBRA. 

in  the  other  that  they  continue  equal,  and  each  approaches  a 
limit,  then  their  limits  are  equal. 

Let  X  and  y  be  the  variables,  a  and  h  their  respective 
limits.     To  prove  a  =  h.     We  have  (§  423) 

a  =  x-^x\  h  =  y-\-y\ 

where  x^  and  y'  are  variables  which  approach  0  as  a  limit. 

Then,  since  the  equation  x^=y  always  holds,  we  have, 
by  subtraction,  a—h  =  x'  —  y\ 

x^  —  y'  can  be  made  less  than  any  assigned  value  since 
x'  and  y'  can  each  be  made  less  than  any  assigned  value. 

Since  x'  —  y'  is  always  equal  to  the  constant  a  —  h, 
x'  —  y'  must  be  a  constant.  But  the  only  constant  which 
is  less  than  any  assigned  value  is  0.  Therefore  x'  —  y'  =  0, 
and  hence  a— 5  =  0.     .',  a  =  b. 

431.  Limit  of  a  Sum.  The  limit  of  the  algebraic  sum  of 
any  finite  number  of  variables  is  the  algebraic  sum  of  their 
respective  limits. 

Let  X,  y,  z, ,  be  variables ; 

a,  b,  c, ,  their  respective  limits.- 

Then  a  —  x,  b—y,  c~z,  ,  are  variables  which  can 

each  be  made  less  than  any  assigned  value  (§  423). 

Then  {a  —  x) -\- {b  —  y) -{- {c  —  z) -\- can  be  made  less 

than  any  assigned  value. 

For,  let  V  be  the  numerically  greatest  of  the  variables  a  —  x,  h  —  y, 
c  —  z ,  and  n  the  number  of  variables. 

Then,  (a  -  re)  +  (6  —  y)  +  (c  —  z)  + <v  +  v  +  v to  n  terms 

<  wv; 

but  nv  can  be  made  less  than  any  assigned  value  since  n  is  finite 
and  V  can  be  made  less  than  any  assigned  value  (^  427,  I.). 

Therefore,  (a  —  cc)  +  (6  —  y)  +  (c  —  z) ,  which  is  less  than  nv,  can 

be  made  less  than  any  assigned  value. 


VARIABLES    AND    LIMITS.  409 

.*.  (a-fS-fc-j- )  — (^'  +  y+2;  + )    can    be    made 

less  than  any  assigned  value. 

.-.  a  +  ^  +  c  + is  the  limit  of  (.r  +  y  +  2  + ).  §  423 

432.  Limit  of  a  Product.  The  limit  of  the  product  of  two 
or  more  variables  is  the  product  of  their  respective  limits. 

Let  X  and  y  be  variables,  a  and  h  their  respective  limits. 

To  prove  that  ah  is  the  limit  of  xy. 

Put  x=^a  —  x\  y  =  h~y^\  then  x^  and  y'  are  variables 
which  can  be  made  less  than  any  assigned  value  (§  423). 

Now,  xy  =  {a  —  x')(b  —  y') 

=  ah  —  ay'  —  hx'  +  x'y\ 
:.  ah  —  xy=^  ay'  +  hx'  —  x'y'. 

Since  every  term  on  the  right  contains  x'  or  y\  the  whole 
right  member  can  be  made  less  than  any  assigned  value 
(§  427,  I.).  Hence,  ah—xy  can  be  made  less  than  any 
assigned  value. 

.*.  ah  is  the  limit  of  xy  (§  423). 

Similarly  for  three  or  more  variables, 

433.  Limit  of  a  Quotient,  The  limit  of  the  quotient  of 
two  variables  is  the  quotient  of  their  limits. 

Let  X  and  y  be  variables,  a  and  h  their  respective  limits. 
Put   a  —  x  =  x',   and   b~y  =  y';    then   x'   and   y'   are 
variables  with  limit  0  (§  423). 

We  have     x==  a  —  x',  y  =  b~  ?/',  and  -  = :• 

TVT  a      x      a      a  —  x'      hx'  —  ay' 

Now  T--  =  T  — 7 -t^TK t: 

h     y      h      b-y'      b{h-y') 

The  numerator  of  the  last  expression  approaches  0  as  a 
limit,  and  the  denominator  approaches  b"^ ;  hence,  the  ex- 
pression approaches  0  as  a  limit  (§  427,  I.). 

.-.?  —  -  approaches  0  as  a  limit.     .*.  7  is  the  limit  of  -• 
b     y   ^^  h  y 


410  ALGEBRA. 

434.  Vanishing  Practions.  When  one  or  more  variables 
are  involved  in  both  numerator  and  denominator  of  a  frac- 
tion, it  may  happen  that  for  certain  values  of  the  variables 
both  numerator  and  denominator  of  the  fraction  vanish. 

The  fraction  then  assumes  the  form  -,  which  is  a  form 

without  meaning ;  as  even  the  interpretation  of  §  428  fails, 
since  the  numerator  is  0.     If,  however,  there  is  but  one 
variable  involved,  we  may  obtain  a  value  as  follows : 
Let  X  be  the  variable,  and  a  the  value  of  x  for  which  the 

fraction  assumes  the  form  -.     Give  to  a;  a  value  a  little 

greater  than  a,  as  a  +  2  I  the  fraction  will  now  have  a  defi- 
nite value.  Find  the  limit  of  this  last  value  as  z  is  indefi- 
nitely decreased.  This  limit  is  called  the  limiting  value  of 
the  fraction. 

(1)  Find  the  limiting  value  of as  x  approaches  a. 

X  —  a 

When  X  has  the  value  a,  the  fraction  assumes  the  form  — 

0 
Put  x  =  a  +  Z]  the  fraction  becomes 

(g  +  2)^  -  z^  _  2az  +  2' 
{a  +  z)  —  z  z 

Since  z  is  not  0,  we  can  divide  by  z  and  obtain  2a-\-  z. 
As  z  is  indefinitely  decreased,  this   approaches   2  a   as  a  limit. 
Hence  2  a  is  the  answer  required. 

(2)  Find  the  limiting  value  of        .^  ~    — ^i—    when  x 

o  X'  -\-  2iX  —  i 

becomes  infinite. 

2a;^  —  4a;  +  5  x^      x^ 


We  have 


3a:3  + 2x^-1      ^^2       1 


X 


7? 


As  X  increases  indefinitely,  -  approaches  0,  and  the  fraction  ap- 
proaches -• 


VARIABLES   AND    LIMITS.  411 

Exercise  126. 
Find  the  limiting  values  of : 

(4ar'  — 3)(1  —  2x)  ^-^en  x  becomes  infinitesimal. 

7a;'-6:r  +  4 

{x  —  b)\x  +  7)  ^]^gjj  ^  becomes  infinite. 

a;* +  35 

(^  +  ^),  ^iien  a:  becomes  infinitesimal. 
a;^  +  4 

rf'  — 807+15  ^j^^^  ^  approaches  3. 

5     X  —  d —   when  X  approaches  —  3. 

0:^  +  9:^+18 

x{x}A-^x-^6)     ^\^Q^  ^  approaches  —  1. 
a;^  +  3o;^  +  5a;  +  3 

7     y  +  o:  —  ^ when  x  approaches  1. 

a;'  +  2a;^-2a;-l 

8,  4^_+V^^_^  when  o:  approaches  1. 
2a;-V.^+i 

9.    ^~  ■ — -—^  when  a;  approaches  1. 

■Vx'-l  +  ^x-l 

10  ^  ~^  when  x  approaches  2. 

Va;  +  2-V3o:~2 

11  V^  —  g  +  V:<7  —  V ^  ^Yien  x  approaches  a. 

■\/x'  -  a' 

12.  If  X  approaches  a  as  a  limit,  and  n  is  a  positive 
integer,  show  that  the  limit  of  of  is  a**. 

13.  If  X  approaches  a  as  a  limit,  and  a  is  not  0,  show 
that  the  limit  of  a;"  is  oT,  where  w  is  a  negative  integer. 


CHAPTER  XXXIV. 

SERIES. 

435.  Oonvergency  of  Series.  For  an  infinite  series  to  be 
convergent  (§  325)  it  is  necessary  and  sufficient  that  the 
sum  of  all  the  terms  after  the  nth,  as  n  is  indefinitely  in- 
creased, should  approach  0  as  a  limit. 

Although  each  of  the  terms  after  the  nth  may  approach  0  as  a 
limit,  their  sum  may  not  approach  0  as  a  limit. 
Thus,  take  the  harmonical  series, 

J     1^    1^    1^  1^    _J_        1 

'    2    3*    4'   n    n  +  l     n  +  2 

Each  term  after  the  nih  approaches  0  as  n  increases. 
The  sum  of  n  terms  after  the  nth  term  is 

/.^-^  +  -L-  +  ^-  + +  -L, 

n  +  1      n  +  2     ?i  +  3  2n 

which  is  > 1 h to  n  terms  ;  therefore  >  n  X  —  ;  that  is,  >  — 

2n     2n  2n  2 

Now,  the  first  term  is  1,  the  second  term  is  J,  the  sum  of  the  next 
two  terms  is  greater  than  ^,  the  sum  of  the  succeeding  four  terms  is 
greater  than  ^ ;  and  so  on.  So  that,  by  increasing  n  indefinitely,  the 
sum  will  become  greater  than  any  finite  multiple  of  J. 

Therefore,  the  series  is  divergent. 

To  determine  whether  the  following  series  is  convergent : 


1  '  [2  '  [3  '        |n-l     [n  '  \n  +  l  ' 

The  rith  term  is  .     The  sum  of  the  remaining  terms  is 

|n  —  1 

i  +  _J_  ._i_  + -l/l   I      1      ,  1  I 

\n     \n  +  l      |n  +  2  [n\        n  +  1      (n  +  l)(n  +  2) 


CONVERGENCY    OF   SERIES.  413 


This  is<  —  (1-^-+  —  + )  ;   therefore,  since  1  +  -  -f  — h 

\n\        n     v?  J  n      n^ 


is  the  expansion  of , 


l/_i_\    or   ^  I     ^'    \ 


n 

1 


that  is,  <  - 


But  as  n  increases  indefinitely,  this  last  expression  approaches  0  as 
a  limit.     Hence,  the  series  is  convergent. 

436.  Test  for  Oonvergency  of  a  Series.  If  the  terms  of  an 
infinite  series  are  all  positive,  and  the  limit  of  the  nth  term 
is  0,  then  if  the  limit  of  the  ratio  of  the  (n  + 1)^*^  terTn  to 
the  nth  terTn,  as  n  is  indefinitely  increased,  is  less  than  1, 
the  series  is  convergent. 

Let  Ui,  U2,  U3, Un,  Un+i,  w„+2i ^6  an  infinite  series. 

Let  r  represent  the  limit  of  the  ratio  -^^  as  n  increases 

indefinitely,  and  suppose  r  to  be  positive  and  less  than  1. 
Let  k  be  some  fixed  number  between  r  and  1,  and  take 

k  so  near  1  that  !^,    ^^:^, ,  shall  each  be  <  k. 


u„      u 


n+l 

Then,  — -<^,       ^r"^'^'       ^;r~<^' 

'.    W„^.i  <  hUn,  ^,1+2  <  kUnj^i,       W„^.3  <  TcUn+2,   

•••  w„+i  +  w„+.,  +  w„+3  + <Un{h-\-  ¥  +  F  + ) 

■*•    W„4.i  +  2<„+2  +  ^n+3  + <  ^«  ..    _   7  > 

since         k-^-V^  ^W  -\- is  the  expansion  of -• 

But,  by  hypothesis,  w„  approaches  0  as  a  limit  as  n  is 
indefinitely  increased.     Hence,  the  series  is  convergent. 
Similarly,  when  r  is  negative,  and  between  0  and  —  1. 


414  ALGEBRA. 

Thus,  in  the  series 

1.14-1  +  1  + ^+1 

1      \2     [3  \n-l      \n 

,  and  this 
Un         n 


^'^^  =  -,  and  this  approaches  0  as  a  limit  as  n  is  indefinitely  in- 


creased ;  moreover,  the  nth  term,  ,  approaches  0  as  a  limit. 

\n  —  1 

Hence,  the  series  is  convergent. 

If  r  >  1,  there  must  be  in  the  series  some  term  from 
which  the  succeeding  term  is  greater  than  the  next  preced- 
ing term ;  so  that  the  remaining  terms  will  form  an  in- 
creasing series,  and  therefore  the  series  is  not  convergent. 

If  r  =  rb  1,  this  value  gives  no  information  as  to  whether 
the  series  is  convergent  or  not;  and  in  such  cases  other 
tests  must  be  applied. 

If  r  <  1,  but  approaches  1,  or  —  1,  as  a  limit,  then  no 
fixed  value  k  can  be  found  which  will  always  lie  between 
r  and  ±  1,  and  other  tests  of  convergency  must  be  applied. 

Thus,  in  the  infinite  series 

1 ■ 1+1+ +1+    1     ■ 


im      2"*      3"*  n~      (n  +  1)"» 

r,  the  ratio  of  the  {n  +  \)th  term  to  the  nth  term,  is 

which  approaches  1  as  a  limit  as  n  increases. 

Suppose  m  positive  and  greater  than  1 ;  then  the  first  term  of  the 

9 

series  is  1.     The  sum  of  the  next  two  terms  is  less  than  -^.    The  sum 
of  the  next  four  terms  is  less  than  — .     The  sum  of  the  next  eight 

Q 

terms  is  less  than  —  ;  and  so  on.  Hence,  the  sum  of  the  series  is  less 

than    l+A+A  +  -i  + or  <  1  +  J_  +  _i_  +  J-  + , 

2"*     4m     8"*  2"*-!      4"*-i      8"*-! 

which  is  evidently  convergent  when  m  is  positive  and  greater  than  1. 


CONVERGENCY    OF   SERIES.  415 

If  m  is  positive  and  equal  to  1,  the  given  series  becomes 

l+^  +  i  +  i  + 

which  is  the  harmonical  series  shown  in  §  435  to  be  divergent. 

If  m  is  negative,  or  less  than  1,  each  term  of  the  series  is  then 
greater  than  the  corresponding  term  in  the  harmonical  series,  and 
hence  the  series  is  divergent. 

437.  Special  Case.  If  the  terms  of  an  infinite  series  are 
alternately  positive  and  negative;  if,  also,  the  terms  contin- 
ually decrease,  and  the  limit  of  the  nth  term  is  zero,  then 
the  series  is  convergent. 

Consider  the  infinite  series, 

U1  —  U2  +  Us  —  u^-\r =F  i^„  =h  Un^x  =+=  w„+2  d= 

The  sum  of  the  terms  after  the  nth  term  is 

which  may  be  written 

Since  the  terms  are  continually  diminishing,  each  of  the 
groups  in  either  form  of  expression  is  positive,  and  there- 
fore the  absolute  value  of  the  required  sum  is  seen,  from 
the  first  form  of  expression,  to  be  less  than  u^i^x ;  and  from 
the  second  form  of  expression,  to  be  greater  than  Wn+i— Wn+a- 
But  both  w„+i  and  w„+2  approach  zero  as  n  increases  indefi- 
nitely ;  therefore  the  sum  of  the  series  after  the  nth  term 
approaches  zero,  and  the  series  is  convergent. 

In  finding  the  snra  of  an  infinite  decreasing  series  of  which  the 
terms  are  alternately  positive  and  negative,  if  we  stop  at  any  term, 
the  error  will  be  less  than  the  next  succeeding  term. 

The  series      1--  +  --T  + ±  -  =f ± is  convergent. 

2     3     4  n    n+1 


lU. 


416  ALGEBRA. 

For,  we  may  write  the  series 

i-i+a-i)  +  a-i)  + .ori-a-i)-a-i)- , 

which  shows  that  its  sum  is  greater  than  ^,  and  less  than  1. 

Observe  that  the  present  test  applies  to  series  in  which 
^^*  approaches  1,  or  —  1,  as  a  limit.  To  such  series  the 
test  of  §  436  will  not  apply. 

438,  Oonvergency  of  the  Binomial  Series.  In  the  expan- 
sion of  (1  +  xy,  the  ratio  of  the  (r  -f  l)th  term  to  the  rth 
term  is  (§  340) 

n  —  r-i-1  /n  +  1 

-^—x,  or    — ^ 

r  \    0' 

If  X  is  positive  and  r  greater  than  n-}-!,  the  expression 
^^^-i^ 1  is  negative ;  hence  the  terms  in  which  r  is  greater 

than  w  + 1  are  alternately  positive  and  negative. 

If  X  is  negative,  the  terms  in  which  r  is  greater  than 
n-\-l  are  all  positive.     In  either  case  we  have 

Ur       \     r  J 

as  r  is  indefinitely  increased,  this  approaches  the  limit  —  x. 
Hence  (§  436),  the  series  is  convergent  if  x  is  numerically 
less  than  1. 

If  n  is  fractional  or  negative,  the  expansion  of  (a  +  hy  must  be 
in  the  form  a4l  +  -]"  if  a>h;   and  in  the  form  Wl  +  -\     if 


h  >  a  (g  344). 

439. 

Examples. 

(1) 

For  what  values  of  x  is 

the 

infinite  series 

2^3                n 

convergent? 

CONVERGENCY    OF   SERIES.  417 

Here,  r  =  -^^—  =  I )  a;  =  (  1 ]  x. 

Un       \n  +  lj         \       n  +  ij 

As  n  is  indefinitely  inereased,  r  approaches  a;  as  a  limit.  Hence, 
the  series  is  convergent  when  x  is  numerically  less  than  1 ;  and 
divergent  when  x  is  numerically  greater  than  1. 

When  a;  =  1,  the  series  is  convergent  by  §  437. 

When  x=  —  1,  the  series  becomes 

-fi+Ui +i  +  .....V     . 

V        2      3  n  j 

the  harmonical  series  already  shown  to  be  divergent  (§  435). 

(2)  For  what  values  of  x  is  the  infinite  series 

cr  01^  X  x*^ 

-i f- -convergent? 


1x22x33x4  71(71+1) 


Here.  r  =  -^^ 


(-)"(TT1)^ 


Un 

As  n  is  indefinitely  increased,  r  approaches  a;  as  a  limit. 
If  a;  is  numerically  less  than  1,  the  series  is  convergent. 
If  a;  is  numerically  greater  than  1,  the  series  ia  divergent. 
If  a:  =  1,  every  term  of  the  series 

1  +^  +  ^  + 


1x2     2x3     3x4 
is  less  than  the  corresponding  term  of  the  series 

'^hh 

This  last  series  is  a  special  case  of  the  series 

■  i+J-+i-+ 

im        2"*        3'» 

and  is  therefore  convergent  (g  436). 

Hence,  \- 1 + is  convergent. 

1X2      2x3      3x4  ^ 

If  a;  =  —  1,  the  series  becomes 


-1-  +  ^ L_ 

1x2     2x3     3x4 


and  is  convergent  by  §  437. 


418  ALGEBRA. 

Series  of  Differences. 

440.  Definitions.  If,  in  any  series,  we  subtract  from  each 
term  the  preceding  term,  we  obtain  a  first  series  of  differ- 
ences; in  like  manner  fiom  this  last  series  we  may  obtain  a 
second  series  of  differences  ;  and  so  on.  In  an  arithmetical 
series  the  second  differences  all  vanish. 

There  are  series,  allied  to  arithmetical  series,  in  which 
not  the  first,  but  the  second,  or  third,  etc.,  differences  vanish. 

Thus  take  the  series 

1        5        12        24        43        71        110 
1st  differences,  4         7        12        19        28        39 

2d  differences,  3         5  7  9         11 

3d  differences,  2         2  2  2 

4th  differences,  0  0  0 

In  general,  if  «!,  aj,  «3,  he  such  a  series, 

^1,  ^2,  ^3,  be  the  first  differences, 

Ci,  C2,  c-i,  be  the  second  differences, 

c?i,  c?2,  d-s,  be  the  third  differences, 

61,  62,  63,  be  the  fourth  differences, 

we  have  rti       ^2       (^3       «*       ci^      «6       «7 

1st  differences,  b^       b.^       b^       bi       b^       b^ 

2d  differences,  Cy       c^       c^        c^       c^     

3d  differences,  d^       d.^       d-i       d^ 

4th  differences,  e^       62       e^ 

and  finally  arrive  at  differences  which  all  vanish. 

441.  Any  Required  Term.  Let  us  take  a  series  in  which 
the  fifth  series  of  differences  vanishes.  Any  other  case  can 
be  treated  in  a  manner  precisely  similar.  From  the  way 
in  which  the  successive  series  are  formed,  we  have : 


SERIES    OF   DIFFERENCES.  419 

«2  =  «i  +  ^1  «3  =  «2  +  i>2  =  «i  +  2  5,  +  (?i 

^2  =  ^1  +  ^1  h^  =  h.,-\-c<,  =  h^  +  2c^-\-d^ 

^2  =  ^1  +  <^i  6?3  =  ^2  +  c?2  =  Ci  +  2c?i-f  ei 

c?2  =  <^i+  ^1  ^3  =  c?24-  ^2  =  c?i+  2ei 

^4  =  ^3  +  ^3  =  ^1  +  3  <?i  +  3  c/i  +  61 
^4  =  ^3  +  <^3=  ^1  +  3  (ii  +  3  ei 

C?4=C?3+e3  =  (^1+3^1 

«5  =  «4  +  ^4  =  «1  +  4  ^1  +  6  .^1  +  4  C?i  +  ^1 

^5  =  ^4  +  ^4  =  ^1  +  4^, +  6c?,  +  4ei 

^5   =  ^4  +  <^4=  ^1  +  4  C?i  +  6  ^1 

«6=«5+^5=«l+  5^1+  10^1  +  10(fi  +  5ei 

^6=^5+C5  =  ^l+5^1  +  10c?i+10^1 

«7  =  «6  +  ^6  =  «i  +  6  5i  4- 15  (?i  +  20  c7i  +  15  ex 
and  so  on. 

The  student  will  observe  that  the  coefficients  in  the  ex- 
pression for  a^  are  those  of  the  expansion  of  {x  +  y)\  and 
similarly  for  a^  and  a^ ;  hence,  in  general,  if  we  represent 
«!,  ^1,  ^1,  etc.,  by  a,  6,  r?,  etc.,  we  have,  putting  for  the 
(n  +  l)th  term  a„+i,  the  formula 

„„,..„+„,  +  tfcLW«J.L^^^.+ 

Ex.    Find  the  11th  term  of  1,  5,  12,  24,  43,  71,  110, 

Here  (g  440)  a  =  1,  6  -  4,  c  =  3,  (?=  2,  e  =  0 ;  and  n  -  10. 
.-.  aii  =  a +  106 +  45c  + 120c? 

-  1  +  40  +  135  +  240  =  416. 


420  ALGEBRA. 

442.   Sum  of  the  Series.     Form  a  new  series  of  which  the 

first  term  is  0,  and  the  first  series  of  differences  ai,  a2,  a^, 

This  series  will  be  the  following : 

0,  au  ai  +  «2,  ai  +  «2  +  «3,  ai  +  «2  +  «3  +  «4,  

The  (n  +  l)th  term  of  this  series  will  be  the  sum  of  n 

terms  of  the  series  ai,  a.2,  a^, 

Find  the  sum  of  11  terms  of  the  series  1,  5,  12,  24,  43,  71, 

The  new  series  is        0        1        6        18        42        85        156 
First  differences,  1        5        12        24        43        71 

Second  differences,  4        7        12        19        28 

Third  differences,  3         5  7  9 

Fourth  differences,  2         2  2 

Here  a  =  0,  6  =  1,  c  =  4,  cZ  =  3,  e  =  2;  and  n  =  11. 
.-.  s  =  a  +  116  +  55c  +  165^  +  330e 
=  11  +  220  +  495  +  660 
=  1386. 

If  s  is  the  sum  of  n  terms  of  the  series  «!,  aj,  «3, 


f.  ,          .  n(n  —  l)  J    .  n(n—l)(n  —  2)      , 
1x2       ^       1x2x3         ^ 
Ex.    Find  the  sum  of  the  squares  of  the  first  n  natural 
numbers,  l^  2^  3^  4^ n\ 

Given  series,  1        4        9        16        25      n"^ 

First  differences,  3        5        7  9 

Second  differences,  2        2         2 

Third  differences,  0        0 

Therefore,  a  =  1,  5  =  3,  c  =  2,  cZ  =  0. 
These  values  substituted  in  the  general  formula  give 

s  =  n  +  ^(^-^)x3+^^^-^X^-^)x2 
1X2  1x2x3 

=  ^(6  +  9n-9  +  2n2-6n  +  4} 
6 

=  ^(2n'' +  3  n  +  n  =  !L(!L±i)(2n+i). 
6^  ^  6 


SERIES    OF    DIFFERENCES.  421 

443.  Piles  of  Spherical  Shot.  I.  When  the  pile  is  in  the 
form  of  a  triangular  pyramid,  the  summit  consists  of  a 
single  shot  resting  on  three  below ;  and  these  three  rest 
on  a  course  of  six;  and  these  six  on  a  course  of  ten,  and 
so  on,  so  that  the  courses  will  form  the  series, 

1,  1  +  2,  1+2  +  3,  1  +  2  +  3  +  4 1  +  2+ +  7i. 


Given  series,  1         3        6        10        15 
First  differences,  2         3        4  5 

Second  differences,  111 

Third  differences,  0        0 

Here,  a-1,  &  =  2,  c  =  1,  d=^0. 

These  values  substituted  in  the  general  formula  give 

2  2x3 


n<l+n  — 


1  + 


w2-3w  +  2 


=  ^{(n  +  l)(n  +  2)} 

_n(n  +  l){n  +  2) 
1x2x3 

in  which  n  is  the  number  of  balls  in  the  side  of  the  bottom  course,  or 
the  number  of  courses. 

II.  When  the  pile  is  in  the  form  of  a  pyramid  with  a 
square  base,  the  summit  consists  of  one  shot,  the  next  course 
consists  of  four  balls,  the  next  of  nine,  and  so  on.  The 
number  of  shot,  therefore,  is  the  sum  of  the  series, 

r,  2^  3^  4^ ^.i 

Which,  by  ?  442,  is 

n(n  +  l)(2n  +  1) 
1X2X3 

in  which  n  is  the  number  of  balls  in  the  side  of  the  bottom  course,  or 
the  number  of  courses. 


422  ALGEBRA. 

III.  When  the  pile  has  a  base  which  is  rectangular,  but 
not  square,  the  pile  will  terminate  with  a  single  row.  Sup- 
pose j9  the  number  of  shot  in  this  row ;  then  the  second 
course  will  consist  of  2(^  +  1)  shot;  the  third  course  of 
3(p  +  2);  and  the  nth  course  of  n{p-\-n—V).  Hence 
the  series  will  be 

p,  2p  +  2,  3^  +  6,  ,  n{p  +  n-l). 

Given  series,  p      2p  +  2      3p  +  Q      4p  + 12 

First  differences,  p  +  2        p  +  4        p  +  Q 

Second  differences,  2  2 

Third  differences,  0 

Here,     a^p,  6  =p  +  2,  e  =  2,  d=0. 

These  values  substituted  in  the  general  formula  give 

,  n(n  —  l)  ,      .  o\  ■  w(^  —  ly^  —  2) , ,  o 
8=np  +  -L-_^  {p  +  2)+    \^l''^^     '  X  2. 

•=  7  {6i>  +  3(n  -  \){p  +  2)  +  2(n  -  l)(n  -  2)} 


=  -(6j9  +  3np-3j9  +  6n-G  +  2n2-6n  +  4) 
6 


=  -(3np  +  3p  +  2n2-2) 


-^(n+l)(3p  +  2n-2). 
b 

If  n^  denote  the  number  in  the  longest  row,  then  n^  =p  +  n  —  1, 
and  therefore  p  =  n^  —  n  +  1;  and  the  formula  may  be  written. 

«^^(n  +  l)(3n^-n  +  l), 
o 

in  which  n  denotes  the  number  in  the  width,  and  n^  in  the  length, 
of  the  bottom  course. 

"When  the  pile  is  incomplete,  compute  the  number  in  the 
pile  as  if  complete,  then  the  number  in  that  part  of  the  pile 
which  is  lacking,  and  take  the  difference  of  the  results. 


SERIES   OF    DIFFERENCES.  423 


Exercise  127. 


1.  Find  the  fiftieth  term  of  1,  3,  8,  20,  43, 

2.  Find  the  sum  of  the  series  4,  12,  29,  55, to  20  terms. 

3.  Find  the  twelfth  term  of  4,  11,  28,  55,  92,  

4.  Find  the  sum  of  the  series  43,  27,  14,  4,  —  3,  to 

12  terms. 

5.  Find  the  .seventh  term  of  1,  1.235,  1.471,  1.708, 

6.  Find  the  sum  of  the  series  70,  66,  62.3,  58.9,  to 

15  terms. 

7.  Find  the  eleventh  term  of  343,  337,  326,  310, 

8.  Find  the  sum  of  the  series  7  X  13,  6  X  11,  5  X  9,  

to  9  terms. 

9.  Find  the  sum  of  n  terms  of  the  series  3x8,  6x11, 

9x14,  12X17, 

10.  Find  the  sum  of  n  terms  of  the  series  1,  6,  15,  28,  45, 

11.  Determine  the  number  of  shot  in  the  side  of  the  base 

of  a  triangular  pile  which  contains  286  shot. 

12.  The  number  of  shot  in  the  upper  course  of  a  square 

pile  is  169,  and  in  the  lowest  course  1089.     How 
many  shot  are  there  in  the  pile  ? 

13.  Find  the  number  of  shot  in  a  rectangular  pile  having 

17  shot  in  one  side  of  the  base  and  42  in  the  other. 

14.  Find  the  number  of  shot  in  five  courses  of  an  incomplete 

triangular  pile  which  has  15  in  one  side  of  the  base. 

15.  The  number  of  shot  in  a  triangular  pile  is  to  the  num- 

ber in  a  square  pile,  of  the  same  number  of  courses, 
as  22  :  41.     Find  the  number  of  shot  in  each  pile. 

16.  Find  the  number  of  shot  required  to  complete  a  rec- 

tangular pile  having  15  and  6  shot,  respectively,  in 
the  sides  of  its  upper  course. 


424  ALGEBRA. 

17.  How  many  shot  must  there  be  in  the  lowest  course 

of  a  triangular  pile  so  that  10  courses  of  the  pile, 
beginning  at  the  base,  may  contain  37,020  shot? 

18.  Find  the  number  of  shot  in  a  complete  rectangular 

pile  of  15  courses  which  has  20  shot  in  the  longest 
side  of  its  base. 

19.  Find  the  number  of  shot  in  the  bottom  row  of  a  square 

pile  which  contains  2600  more  shot  than  a  triangular 
pile  of  the  same  number  of  courses. 

20.  Find  the  number  of  shot  in  a  complete  square  pile  in 

which  the  number  of  shot  in  the  base  and  the  num- 
ber in  the  fifth  course  above  differ  by  225. 

21.  Find  the  number  of  shot  in  a  rectangular  pile  which 

has  600  in  the  lowest  course  and  11  in  the  top  row. 

Interpolation. 

444.    As  the  expansion  of  {a  -\-  hy  has  the  same  form  for 
fractional  as  for  integral  values  of  n,  the  formula 

,      1    .  n(n  —  V)      .  n(n—-V)(n  —  2)  J  . 

may  be  extended  to  cases  in  which  w  is  a  fraction,  and  be 
used  to  interpolate  terms  in  a  series  between  given  terms. 

(1)  The  cube  roots  of  27,  28,  29,  30,  are  3,  3.03659, 
3.07232,  3.10723.     Find  the  cube  root  of  27.9. 

3        3.03659        3.07232        3.10723 
First  differences,  0.03659        0.03573        0.03491 

Second  differences,  -  0.00086       -  0.00082 

Third  differences,  0.00004 

These  values  substituted  in  the  general  formula  give 

-  3  +  0.032931  +  0.0000387  +  0.00000066  =  3.03297. 


SERIES   OF    DIFFERENCES.  425 

(2)  Given  log  127  -  2.1038,  log  128  =  2.1072,  log  129  = 
2.1106.     Find  log  127.37. 

0.0034        0.0034  =  first  order  of  differences. 

0  -=  second  order  of  differences. 

Therefore,  the  differences  of  the  second  order  will  vanish,  and 
the  required  logarithm  will  be 

2.1038  +  j%\  of  0.0034 
=  2.1038  +  0.001258 
=  2.1051. 

(3)  The  latitude  of  the  moon  on  a  certain  Monday  at 
noon  was  1°  53'  18.9",  at  midnight  2°  27'  8.6" ;  on  Tuesday 
at  noon  2°  58'  55.2",  at  midnight  3°  28'  5.8"  ;  on  Wednes- 
day at  noon  3°  54' 8.8".  Find  its  latitude  at  9  p.m.  on 
Monday. 

The  series  expressed  in  seconds,  and  the  differences,  will  be 

6798.9        8828.6        10735.2        12485.8        14048.8 

2029.7        1906.6  1750.6  1563 

-123.1         -156         -187.6 

-32.9         -31.6 

1.3 

As  9  hours  =  |  of  12  hours,  n  =  |. 

Also,  a  =  6798.9,   6  =  2029.7,   c  =  - 123.1,   (f  =  -32.9,   e-1.3. 

These  values  substituted  in  the  general  formula 

1X2  1x2x3 

n{n-l)(7i-2(n-3)^ 
1x2x3x4 


give  6798.9  + 1(2029.7)  - 1(-  ^)( 


i23.r 

2    . 


+ 


=  6798.9  +  1522.27  +  11.54  -  1.29 
=  8331.4  =  2°  IS^  51.4^^ 


426  ALGEBRA. 


COMPOUND    SERIES. 

445.    It  is  evident,  from  tlie  form  of  certain  series  that 
they  are  the  sum  or  the  difference  of  two  other  series. 

(1)  Find  the  sum  of  the  series 

1            1            1  1 

1x2'    2x3'    3x4    n(n-\-l) 

Each  term  of  this  series  may  evidently  be  expressed  in  two  parts  : 
1111  1         1 


12     2     3  n     n+1 

so  that  the  sum  will  be 

in  which  the  second  part  of  each  term,  except  the  last,  is  cancelled 
by  the  first  part  of  the  next  succeeding  term. 

Hence,  the  sum  is  1 

n  +  1 

As  n  increases  without  limit,  this  sum  approaches  1  as  a  limit. 

(2)  Find  the  sum  of  the  series 

111  1 


3x5    4x6     5x7  7i(n  +  2) 

Each  term  may  be  written, 

2\,3     5J     2V4     6/  2^1     n  +  2) 

...S«m  =  l/'i  +  l  +  l  +  l  + +  1-1-1-  1 

2V3     4     5     6 


2\3     456  w56  n     n  +  2 

1/1 

2  , 

7  1  1 


\3      4      n  +  1      n  +  2y 


Hence,  the  sum  is  , 

24      2  (n  +  1)      2  (n  +  2) 

As  n  increases  without  limit,  this  sum  approaches  ^^  as  a  limit. 


COMPOUND   SERIES.  427 

Exercise  128. 

Sum  to  n  terms,  and  to  infinity,  the  following  series : 

1.       1  1  1 


1x4     2x5     3x6 

1  1  1 


1x3x5  2x4x6  3x5x7 
1        1        1 


2x4x6  4x6x8  6x8x10 

4        7       10 

2x3x4'  3x4x5'  4x5x6' 

1  1_ 1 

1x2x3'  2x3x4'  3x4x5' 


446.   Eeversion  of  a  Series,     Given 

y  =^  ax  -\-  b:t^  -\-  cx^  -\-  dx*'  + 

where  the  series  is  convergent,  to  find  x  in  terms  of  y. 

Assume        x  —  Ay  ■\-  By"^  -\-  Cy^  -f-  Dy*-  + 

In  this  series  for  y  put  ax -\- hx^ -\- cx^  +  dx^ -\- ;  then 

X  =  aAx  -{-hA     x"^  -\-  cA        o(?  + 
-\-a^B        -\-2ahB 
+  a'C 

Comparing  coefficients  (§  330), 

aA  =  l)  bA-\-a'B  =  0]  cA  +  2abB  +  a^C=  0. 

,.^^1,   B  =  -\,    C=2^-=^,  etc. 
a  a  a      » 


428  ALGEBRA. 

(1)  Given  y  =  x  -\- x^ -\- x^ -{- ;  find  x  in  terms  of  y. 

Here,  a=\,    6=1,        c  =  1,     d^l,  

^  =  1,    ^=.-1,    (7^1,   Z>  =  -1 

Hence,  x=-y  —  y^  +  y^-y^  + 

(2)  Revert  2/  =  a;-|  +  |-^  + 

Here,  a  =  1,     &  =  -  i,     c  =  i      (^  =  -  ^,  

...  ^  =  1,   5  =  i,      (7=i,    i)  =  i 

[2  (3  [4 

Hence,  .  =  y  +  |  +  |  +  g  + 


Exercise  129. 
Revert : 

1.    y^a:~2x'-{'Zx'-4:x'^ 


jr^  ,  x^      x^  , 

^-  ^="-3  +  5-7+ 


7/'  X  X 

3-  y=-+r2+2^+F4+ 


1  4-:r 

447,   Recurring  Series,     From  the  expression ^ 

J.  —  Ax  —  X 

we  obtain  by  actual  division  the  infinite  series 

l-{.3x+7x'  +  11x'  +  4:lx*-i-99a^+ 

In  this  series  any  required  term  after  the  second  is  found 
by  multiplying  the  term  before  the  required  term  by  2x, 
the  term  before  that  by  x*,  and  adding  the  products. 


RECURRING    SERIES.  429 

Thus,  take  the  fifth  term  : 

ilx'  =  2x(17  x')  +  x\7  x^). 
In  general,  if  Un  represent  the  nth  term, 

A  series  in  which  a  relation  of  this  character  exists  is 
called  a  recurring  series.     Recurring  series  are  of  the  first, 

second,  third, order,  according  as  each  term  is  dependent 

upon  one,  two,  three, preceding  terms. 

A  recurring  series  of  the  first  order  is  evidently  an  ordi- 
nary geometrical  series. 

In  an  arithmetical,  or  geometrical,  series  any  required 
term  can  be  found  when  the  term  immediately  preceding 
is  given.  In  a  series  of  differences,  or  a  recurring  series, 
several  preceding  terms  must  be  given  if  any  required  term 
is  to  be  found. 

The  relation  which  exists  between  the  successive  terms 
is  called  the  identical  relation  of  the  series ;  the  coeflficients 
of  this  relation,  when  all  the  terms  are  transposed  to  the 
left  member,  is  called  the  scale  of  relation  of  the  series. 

Thus,  in  the  series 

l  +  3x  +  7x^  +  17c(^  +  41a;*  +  99ar^  + 

the  identical  relation  is  , 

Un=2xUn-l  +X^Un-2; 

and  the  scale  of  relation  is 

l-2a;-a;2. 

448.  If  the  identical  relation  of  the  series  is  given,  any 
required  term  can  be  found  when  a  sufiicient  number  of 
preceding  terms  are  given. 

Conversely,  the  identical  relation,  can  be  found  when  a 
sufiicient  number  of  terms  are  given. 


430  ALGEBRA. 

(1)  Find  the  identical  relation  of  the  recurring  series 

l  +  4:X  +  Ux'  +  ^9x'+inx'-\-691a^  +  2084:x'i- 

Try  first  a  relation  of  the  second  order. 

Assume  w„  =pxun-i  +  gx'^Un-2' 

Putting  n  =  3,  and,  then,  n  =  4, 
14:  =  4ip  +  q, 
49  =  14p  +  42", 
whence,  P  =  I.    q  =  0. 

This  gives  a  relation  which  does  not  hold  true  for  the  fifth  and 
following  terms. 

Try  next  a  relation  of  the  third  order. 

Assume  Un  =pxun-i  +  qx'^Un~2  +  rxr^Un 

Putting  n  =  4,  then  n  =  5,  then  n  =  6. 

49=    14p  +    iq+      r, 
171=    Adp  +  Uq+    4:r, 
597  =  171p  +  49^  +  14r; 
whence,  1>=3,  q  =  2,  r  =  — 1. 

This  gives  the  relation 

Un=  ZxUn-l  +  2x^1^-2  -  An-S 

which  is  found  to  hold  true  for  the  seventh  term. 
The  scale  of  relation  is  l  —  Sx  —  2x^  +  a^. 

(2)  Find  the  eighth  term  of  the  above  series. 
Here,  Wg  =  3  xu^  +  2  x'^u^  —  x^u^ 

=  3a;(2084a;6)  +  2 x\597 a^)  -  2.^111  as*) 

'  =  7275a;^ 

449.  Sum  of  an  Infinite  Series.  By  the  sum  of  an  infinite 
convergent  numerical  series  is  meant  the  limit  which  the 
sum  of  n  terms  of  the  series  approaches  as  n  is  indefinitely 
increased ;  a  divergent  numerical  series  has  no  true  sum. 

By  the  sum  of  an  infinite  series  of  which  the  successive 
terms  involve  one  or  more  variables  is  meant  the  generating 
function  of  the  series,  that  is,  the  expression  of  which  the 
sei'ies  is  the  expansion. 


HECUERING  SERIES.  431 

The  generating  function  is  a  true  sum  when,  and  only 
when,  the  series  is  convergent. 

The  process  of  finding  the  generating  function  is  called 
summation  of  the  series. 

450.  Sum  of  a  Eecurring  Series.  The  sum  of  a  recurring 
series  can  be  found  by  a  method  analogous  to  that  by  which 
the  sum  of  a  geometrical  series  is  found  (§  314). 

Take,  for  example,  a  recurring  series  of  the  second  order 
in  which  the  identical  relation  is 

or  Uk  —puu-i  —  qUk-2  =  0. 

Let  s  represent  the  sum  of  the  series  ;  then 

S  =  Wl  +      W2  +      ^3  -}- W„_i  +      Wn, 

—  pS  =        —pUi  —pU.,  — —pUn-2  —pUn-i  —  pU„, 

—  qs=  —  qUi—  : qUn-s  —  qu„-2  —  qUn-i  —  q^n- 

Now,  by  the  identical  relation, 

u^—pUi—qui  =  0,  Ui—pUs—qu^^O, Un—pUn-i—qUn^i=0. 

Therefore,  adding  the  above  series, 

^Ui  +  {U2—  pUi)        pU^  +  g  (^n  +  '^n-^ 

\-p-q  ^-p-q 

Observe  that  the  denominator  is  the  scale  of  relation. 

If  the  series  is  infinite  and  convergent,  w„  and  w„_i  each 
approaches  0  as  a  limit,  and  s  approaches  as  a  limit  the 

fraction  ^i  +  (^-^-^^0. 
l-p-q 

If  the  series  is  infinite,  whether  convergent  or  not,  this 
fraction  is  the  generating  function  of  the  series. 


432  ALGEBRA. 

For  a  recurring  series  of  the  third  order  of  which  the 
identical  relation  is 

we  find         .  ^  ^1  +  K  -pui)  +  (^3  -pu,  -  qu{) 
1 —p  —  q  —  r 

1— ^  —  2'— ?• 

Similarly  for  any  recurring  series. 

(1)  Find  the  generating  function  of  the  infinite  recurring 
series 

1  +  4a:  4- 13^^  +  43^'  +  142:r*  + 

By  2  448  the  identical  relation  is  found  to  be 
Wjt  =  3  xuk-i  +  x^uit-i. 

Hence,  s  =  l+4a;  +  13a;2  +  43ar»  +  142a;*  + 

-3a;s=     -3  re -12a;* -39x3 -129  a;*- 


4a;3_    13, 


—  a'*s  =  —      a;*—    ■aar-    ioa;'  — 

Adding,  (1  -  3a;- a;2)s  =  1  +  x, 

l_3a;-a;»' 

(2)  Find  the  generating  funetion  and  the  general  term 
of  the  infinite  recurring  series 

l-lx-x"-  43:27^  -  49a:*  -  307a:'*  ~ 

Here  ujc  =  xuk-i  +  6  x^ujc-i. 

s=l-7a:-    a;2-43a^-49a;*- 

—     a;s=     —    x  +  7a;2+      a;3^43^4^ 

«6x2s=  -6x2  +  42x3+    6x*  + 

^_      l-8x      _  l-8x 

l-x-6x2     (l  +  2x)(l-3x) 


RECURRING   SERIES.  433 

By  I  331  we  find 

l-8a;  ^2  1 

(l  +  2a;)(l-3a;)      1  + 2  a;      l-3»" 
By  the  binomial  theorem  or  by  actual  division, 

— —  =  1  -  2a;  +  22a;2  -  2'ic3  _j. ^  2'-(-  l)'-a;'"  + , 

1  +  2a; 

— - —  =  1  +  3a;  +  32^2  +  33ar»  + 3'' a;*"  + 

l-Zx 

Hence  the  general  term  of  the  given  series  is 

[2^+1  (-1)*-- 3'-]  a;". 

(3)  Find  the  identical  relation  in  the  series 

12  +  2^  +  3^  +  4^^  +  5^  +  6^  +  7^  + 

The  identical  relation  is  found  from  the  equations 
16=    9p+    Aq+    r, 
25  =  16p+    9gr  +  4r, 
36  =  25p  +  165'  +  9r, 
to  be  Wi  =■  3  ujc^i  —  3  ma_2  +  w*-8. 

Exercise  130. 

Find  the  identical  relation  and  generating  function  of: 

1.  l  +  2x-\-7x'  +  23x'+ie>x'  + 

2.  S  +  2x  +  S:^+7x'  +  18x'  + 

Find  the  generating  function  of : 

3.  2  +  Sx  +  5x^  +  9x^-\-11x*  +  SSa^+ 

4.  1-Qx-\-  9x'  +  27a;'  +  54a:*+  189^'  + 

6.    l  +  5x-{-9a^-{-lSx'-\-17x*  +  21a^  + 

6.  l4-a:-7a;'  +  33a;*-130r^  +  499a:«  + 

7.  S  +  6x+Ux'  +  ^Qa^  +  98x'  +  276s^  + 

Find  the  sum  of  n  terms  of : 

8.  2  +  5  +  10  +  17  +  26  +  37  +  50-f  ..." 

9.  P  +  2H3'  +  4'  +  5'  + 


434 


ALGEBRA. 


Exponential  and  Logarithmic  Series. 
451.   Exponential  Series.     By  the  binomial  theorem 

l+-=l  +  wa:X-+       ^.  ^^     ^  X  - 
\        ??/  n  1x2  n^ 

■  nx  {nx  —  V)(nx  —  2)  w  2.  i 
1X2X3  n''^ 

f        l\         f        l\f        2\ 

X[X )       x[x ][x ) 

1  +  ^+        [2        ^  [3  + 


(1) 


This  equation  is  true  for  all  real  values  of  x,  but  is  only- 
true  for  values  of  n  numerically  greater  than  1,  since  - 
must  be  numerically  less  than  1  (§  438). 

As  (1)  is  true  for  all  values  of  x,  it  is  true  when  a:  =  1. 


+ 


(2) 


Hence  from  (1)  and  (2), 


a:[  .T ]      n- f  .^^  —  -  )( rr ) 


EXPONENTIAL   AND   LOGAEITHMIC  SERIES. 


436 


This  last  equation  is  true  for  all  values  of  n  numerically- 
greater  than  1.  Take  the  limits  of  the  two  members  as  n 
increases  without  limit.     Then  (§  427) 


^+^+i+i+ 


=^+^+1+1+ ■  (^> 


and  this  is  true  for  all  values  of  x.  It  is  easily  seen  by 
§  436  that  the  second  series  is  convergent  for  all  values  of  x ; 
the  first  series  was  proved  convergent  in  §  435. 

The  sum  of  the  infinite  series  in  parenthesis  is  called  the 
base  of  the  natural  system  of  logarithms,  and  is  generally 
represented  by  e  ;  hence,  by  (3), 


e'  =  l  +  x  +  -t.^±;-{- 


\1  \1 

To  calculate  the  value  of  e  we  proceed  as  follows 
1.000000 


2 

1.000000 

3 

0.500000 

4 

0.166667 

5 

0.041667 

6 

0.008333 

7 

0.001388 

8 

0.000198 

9 

0.000025 

0.000003 
Adding,  e=  2.71828. 

452,   In  A  put  ex  in  place  of  x ;  then 

Put  e"  =  a ;  then  c  =  log^a,  and  e"  =  a*. 
.-.  fl"^l  +  log,a+        12        +        13        "^ 

The  series  in  B  is  known  as  the  exponential  series 
duces  to  A  when  we  put  e  for  a. 


B 


436  ALGEBRA. 

453.   Logarithmic  Series.     In  A  put  e*  =  1  +  y ;  then 
a;  =  loge(l+y),  and  by  A, 

rj^  rj^  rji^ 

Revert  the  series  (§  446),  and  we  obtain 

But              a;  =  !og,(l+y). 
.-.  log.(l+y)  =  y-J  +  ^-^  + 


Similarly  from  B, 


^°g-^^+^)=i4:^(^- 


2  "^3      4"^'" 


The  series  in  D  is  known  as  the  logarithmic  series ;  D  re- 
duces to  0  when  we  put  e  for  a. 

In  0  and  D  y  must  be  between  —  1  and  +  1,  or  be  equal 
to  + 1,  in  order  to  have  the  series  convergent  (§  439,  Ex.  1). 

454.  Modulus.     Comparing  0  and  D  we  obtain 
log«(14-2/)=r-^loge(l  +  y); 

logea 
or,  putting  iVfor  1  +  y, 

log«iV^---i-logeiV: 
log«a 

Hence,  to  change  logarithms  from  the  base  e  to  the  base 

a,  multiply  by =logae;  and  conversely, 

logea 

The  number  by  which  natural  logarithms  must  be  multi- 
plied to  obtain  logarithms  to  the  base  a  is  called  the  modu- 
lus of  the  system  of  logarithms  of  which  a  is  the  base. 

Thus,  the  modulus  of  the  common  system  is  logio^. 


EXPONENTIAL   AND    LOGARITHMIC    SERIES.  437 

455.  Calculation  of  Logarithms.  Since  the  series  in  0  and 
D  are  not  convergent  when  x  is  numerically  greater  than  1, 
they  are  not  adapted  to  the  calculation  of  logarithms  in 
general.     "We  obtain  a  convenient  series  as  follows : 

The  equation 

log,(l  +  y)  =  y-|'  +  ^-J+ (1) 

holds  true  for  all  values  of  y  numerically  less  than  1 ; 
therefore,  if  it  holds  true  for  any  particular  value  of  y,  it 
will  hold  true  when  we  put  —  y  for  y ;  this  gives 

iog,(i-y)  =  -y-5-f-J- (2) 

Subtracting  (2)  from  (1),  since 

log.(l  +  2/)  -  Iog.(l  -  y)  =  log,  (j^y 

find    ,og.(L±|)^.(,+f  +  ^i+ ). 


we 


Put      y=-^;  then  i+^^^il, 
and  loge  f-y-j  =  log,(z  +  1)  -log.j 


\2z  +  l 


1  .  1 


3(2z  +  l)'     5(2z+l)' 


+ 


This  series  is  convergent  for  all  positive  values  of  z. 
Logarithms  to  any  base  a  can  be  calculated  by  the  corre- 
sponding series  obtained  from  D ;  viz. : 

loga(z  +  l)  — logaZ 

log.a  V2z  +  1      3(2z  +  !)» ^  5(22  +  1)*^      J 


438  ALGEBRA. 

(1)  Calculate  to  six  places  of  decimals  log^  2,  loge  3,  loge  10, 
logioe. 

In  E  put  2=1;  then  2z  +  1  =  3,  loggZ  =  0, 

and  log,2  =  |  +  -Ar,  +  ^^+     ^      ■ 


3      3x33     5x35      7x3^ 
The  work  may  be  arranged  as  follows : 


2.0000000 


0.6666667  -^  1  =  0.6666667 


0.0740741  ^  3  =  0.0246914 


0.0082305-  5  =  0.0016461 


0.0009145-  7  =  0.0001306 


0.0001016-  9  =  0.0000113 


0.0000113  -J-  11  =  0.0000010 


0.0000013  -J- 13  =  0.0000001 


loge2  =  0.693147 
loge  3  =  loge2  +  f  +  2   ■   2 


5  3  X  53  5  X  55 
=  1.0986123. 
loge  9  =  loge  (32)  =  2  loge  3  =  2.1972246. 

loge  10  =  loge  9  + -I + 


19     3  X  193     5  X  19* 
=  2.1972246  +  0.1053606 
=  2.302585. 

logioc  =--A^=  0.434294. 

loge  10 

Hence,  the  modulus  of  the  common  system  is  0.434294. 
To  ten  places  of  decimals  : 

loge  10  =  2.3025850928, 
logioe  =0.4342944819. 
For  calculating  common  logarithms  we  use  the  series  in  F. 
logio(z  +  l)  — logioz 


0.8685889638 


V22  +  1^3(22- 


3(22  +  1/  '  5(2^+1/ 


EXPONENTIAL    AND   LOGARITHMIC   SERIES.  439 

(2)  Calculate  to  five  places  of  decimals  logioll. 
Put  2  =  10 ;  then  22  +  1  =  21,  log2  =  1. 

logll  =  l  +  0.86S588(i  +  3-^,  +  ^-^,  + ) 


21 
441 


0.868588 


0.041361  ^  1  =  0.041361 
94  ^  3  =  31 


0.041392 

1 


logioll  =  1.04139 

In  calculating  logarithms,  the  accuracy  of  the  work  may  be  tested 
every  time  we  come  to  a  composite  number  by  adding  together  the 
logarithms  of  the  several  factors  (^  428).  In  fact  the  logarithms  of 
composite  numbers  may  be  found  by  addition,  and  then  only  the 
logarithms  of  prime  numbers  need  be  found  by  the  series. 

456.  Limit  of  ( 1  +- )•     By  the  binomial  theorem, 


(-;)"= 


n         1x2         n^ 


n{n~l){n-2)^ 
"^        1X2X3  n'"^ 

=  i+-+ir^-'+^    "A    "^^+ 

This  equation  is  true  for  all  values  of  n  greater  than  x 
(§  438).  Take  the  limit  as  n  increases  without  limit,  x 
remaining  finite ;  then 

limit        /i    ,  x\'_  .    ,  ^  .  a?   ,  a? 


n  infinite  \        n)  !_?.      lA 


limit 
n  infini 


,  fi+^r     §451 

ite\        nj 


440  ALGEBRA. 


Exercise  131. 


Determine  whether  the  following  infinite  series  are  con- 
vergent or  divergent : 

1     1  4-1 4-2.4-2-4- .  ...  3     .?.  4- ^ 4_  ^  4- :^  4_  .  . 

*  2     3'     4'  '    12^2'      3'     4' 

12  93  04  1  Im  Oto  Qm 

[2     [3     [4  ln»^2"^     3"*     4"* 

5.    Show  that  the  infinite  series 

1  1.1  1 


H  + 


1X2     2x2-^3x2=^     4x2^ 
is  convergent,  and  find  its  sum. 

6.  Find  the  limit*  which  Vl  +  ^^  approaches  as  n  ap- 

proaches 0  as  a  limit. 

7.  Prove  that  1  =  2(1  +  1+ 3  + ). 

8.  Calculate  to  four  places,  loge  4,  loge  5,  loge  6,  loge  7. 

9.  Find  to  four  places  the  moduli  of  the  systems  of  which 

the  bases  are  :  2,  3,  4,  5,  6,  7. 

10.  Show  that 

''^\ej     1x2x3  "^3x4x5  "^5x6x7"^ 

11.  Show  that 

log,«_log.5=___  +  _^_j  +  _^__j_t 

12.  Show  that,  if  x  is  positive, 

13.  Show  that  l+|  +  |  +  f+ =  5e. 

L£    [2     li 


CHAPTER  XXXV. 

GENERAL   PROPERTIES   OF   EQUATIONS. 

457.  Functions.  Any  expression  involving  x  is  called  a 
function  of  x.  If  x  is  involved  only  in  powers  and  roots, 
the  expression  is  an  algebraic  function  of  x.  A  rational 
algebraic  function  involves  x  only  in  powers,  not  in  roots. 
A  rational  algebraic  function  of  x  is  integral  if  it  involves 
only  positive  integral  powers  of  x. 

Thus,  3x^   V^-I.  axKb,  V^^,    (^•^^  +  ^^  +  ^)^,  a^^  log(x) 

^a  —  x  dx  +  e 

are  functions  of  x,  the  first  five  being  alsebraic.     ,    - — ^^^^, 

ax'      hx'      ex      d  a;2  +  l'       2a:2  +  3 

— — -^^ ,  are  fractional  rational  functions  of  x.   3  a;^  +  4  a;  —  1, 

ax^  +  bx  +  c,  ex'  —  d,  are  integral  rational  functions  of  x. 

458.  We  shall  hereafter  consider  only  integral  rational 
functions,  called  quantics,  unless  otherwise  expressly  stated. 
The  degree  of  such  a  function  is  the  same  as  the  exponent 
of  the  highest  power  of  x  involved. 

For  brevity  a  function  of  x  is  often  represented  by  f(x), 
F{x),  <f>(x),  or  by  some  similar  notation.  The  value  of  the 
function /(a;)  when  we  put  a  for  x  is  represented  by /(a). 

Thus,  if  f{x)  =  2  a;3  -  3  x2  -  4  a;  +  5, 

^(2)  =  2(2)3  _  3  (2)2  -  4(2)  +  5  =  1. 

459.  Equations,  An  equation  which  involves  only  inte- 
gral rational  functions  of  x  is  called  a  rational  integral 


442  ALGEBRA. 

equation.     Every  such  equation  can  be  reduced,  by  trans- 
posing all  the  terms  to  the  first  member,  to  the  general  form 

ttoa;"  +  ctix""-'  +  «2^""'  + +  a,,  =  0, 

or  briefly,  f{x)  ^  0. 

The  degree  of  the  equation  is  the  same  as  that  of  the 
function /(a;).  An  equation  of  the  first  degree  is  called  a 
Hnear  equation.  Those  of  higher  degrees  are  called  in 
order  quadratic,  cubic,  biquadratic,  quintic,  etc. 

The  roots  of  an  equation  are  those  values  of  x  for  which 
the  function /(a;)  vanishes. 

460.  Fundamental  Theorems,  Theorem  I,  If  the  function 
f(x)  is  divisible  hy  x  — h,  then  h  is  a  root  of  the  equation 
f  (x)  =  0.  For,  if  <^  {x)  is  the  quotient  obtained  by  divid- 
ing/(a;)  by  a;  —  A,  we  have 

f{x)={x-h)^{x\ 
and  the  equation /(a;)  =  0  may  be  written 
{x—h)^{x)=^. 
But  h  is  obviously  a  root  of  this  equation. 

461.  Theorem  II.  Conversely ,  if  h  is  a  root  of  the  equation 
f  (x)  =  0,  then  f  (x)  is  divisible  by  x  —  ln. 

For  example,  consider  the  equation 

f(x)  =  ax^  -j- bx'^ -^  ex -{- d  ^  0, 
Now,  since  A  is  a  root  of  the  equation  f(x)  =  0,  we  have 

0  =  ah'  +  bh'  +  ch  +  d. 
Subtracting, 

f(x)  =  a(x'  -  h')  +  b(x'-  h')  +  c{x  -  h). 

But  every  term  of  the  second  number  is  divisible  by 
x~h,  and  consequently  f{x)  is  also  divisible  by  x  —  h. 
Similarly  for  any  other  equation. 


GENERAL    PROPERTIES    OF   EQUATIONS.  443 

462.   Synthetic  Division.     Let  the  function 

be  divided  by  a;  —  3. 
The  work  is  as  follows  : 

2a^-3x^-5o^+      x^-d3x-7\x-3 


2x5- 

-6x* 

3x^-ba^ 

3x*-9;^ 

4^3+      x^ 

4x3-12x2 

13x2 -33x 

13x2 -39x 

6x-    7 

6x-18 

2x*  +  3x3  +  4x2+13x  +  6 


11 
The  work  may  be  abridged  by  omitting  the  powers  of  x,  and  writ- 
ing only  the  coefficients.     "We  then  have 

2-3-5+    1-33-    7|l-3 

2-6  2  +  3+4  +  13  +  6 

3-5 


3-9 


4+    1 
4-12 


13- 

-33 

13- 

-39 

6- 

■    7 

6- 

-18 

11 

But  the  operation  may  be  still  further  abridged.  As  the  first  term 
of  the  divisor  is  unity,  the  first  term  of  each  remainder  is  the  next 
term  of  the  quotient.  Again,  we  need  not  bring  down  the  several 
terms  of  the  dividend.  Finally,  we  need  not  write  down  the  first 
terms  of  the  partial  products. 


444  ALGEBRA. 

The  operation  is  then  as  follows : 

2-3-5  +  1- 33 -7[l-3 
-6 

3 
-9 


4 

-  12 


13 
-39 


18 


11 

Omitting  the  first  term  of  the  divisor  as  superfluous,  changing  —  3 
to  +  3,  and  adding  instead  of  subtracting,  we  have,  on  raising  the 
terms  and  bringing  down  the  first  coefficient, 

2-3-5+    1-33-    7|3 
6  +  9  +  12  +  39  +  18 

2  +  3  +  4  +  13+    6  +  11 

The  last  term  below  the  line  gives  us  the  remainder,  the  preceding 
terms  the  coefficients  of  the  quotient.  In  this  particular  problem  the 
quotient  is  2x^  +  3ot^  +  4iX^  +  lSx  +  6,  and  the  remainder  is  11. 

This  method  is  called  the  method  of  Synthetic  Division, 
For  the  application  of  this  method  to  the  division  of  any 
quantic  hjx~h  we  have  the  following  rule  : 

Write  the  coefficients  a,  b,  c,  etc.,  in  a  horizontal  line. 

Bring  down  the  first  coefficient  a. 

Multiply  a  5y  h,  and  add  the  product  to\i. 

Multiply  the  sum  so  obtained  by  h,  and  add  the  product 
to  0. 

Continuing  this  process,  the  last  sum  will  be  the  remainder, 
and  the  preceding  sums  the  coefficients  of  the  quotient. 

Remark.  If  there  are  any  powers  of  x  missing,  their  places  are  to 
be  supplied  by  zero  coefficients. 


GENERAL    PROPERTIES    OF   EQUATIONS.  445 

463.  Value  of  a  Quantic.  By  the  principles  of  division  it 
is  evident  that  the  operation  of  dividing  a  given  quantic 
f{x)  hj  x  —  h  can  be  carried  on  until  the  remainder  does 
not  involve  x.  Represent  the  quotient  by  <^(a:),  and  the 
remainder  by  R.     Then  we  have 

f{x)  =  {x-h)<i>{x)-^R. 

Putting  h  for  x, 

/(A)  =  0  +  i?. 

Hence  the  value  which  a  quantic  f  (x)  assumes  when  we 
put  h.  for  X  is  equal  to  the  last  remainder  obtained  in  the 
operation  of  dividing  f  (x)  by  x  —  h. 

Exercise  132. 

Find  the  quotient  and  remainder  obtained  by  dividing 
each  of  the  following  expressions  by  the  divisor  opposite  it. 


1. 

x'-^bx'-lx-^ 

x-2. 

2. 

x'-lx^-\-bx'-l0x+l2 

x-^. 

3. 

o.^*_|.3^3_.g^2_4^_24 

x-2. 

4. 

r'-^a^  +  ^x'-b 

x-^. 

5. 

?>x'-^x'-^lx-lO 

x-^S. 

6. 

x''-{-^x'-\-x  +  ^ 

X  +  -V 

Are  the  following  numbers  roots  of  the  equations  oppo- 
site them  ? 


7. 

(2) 

3^-Sx'-\-^x-{-i  =  0. 

8. 

(-3) 

x*-d:f^-{-7x'-9x  +  84.  =  0. 

9. 

(-5) 

^5^(3^4_l_7^,3_|_9^_5_0. 

.0. 

(0.2) 

.^3-2.20;^  + 3.4a; -0.6  =  0. 

446  ALGEBRA. 

Find  the  values  of  the  following  expressions  when  for  x 
we  put  the  number  opposite  the  expression. 

11.  2a;' +  3a;' +  5a; -10  (2). 

12.  3a;^-6a;*  +  2a;'  +  3a;-15        (3). 

13.  a;*  -  4a;' +  7a;' +  9a; +  12  (-3). 

464,  Number  of  Eoots.  "We  shall  assume  that  every 
rational  integral  equation  has  at  least  one  root.  The  proof 
of  this  truth  is  beyond  the  scope  of  the  present  chapter.* 

Let /(a;)  =  0  be  a  rational  integral  equation  of  the  nth 
degree.  This  equation  has,  by  assumption,  at  least  one 
root.     Let  a^  be  a  root. 

Then,  by  §  460,    f{x)  -  (a;  -  aj/i  (a;), 

where /i (a;)  is  a  quantic  of  degree  n—1. 

The  equation  /i(a:)==0  must,  by  assumption,  have  a 
root.     Let  Uj  be  a  root. 

Then,  by  §  460,  f,{x)  =  (x  -  a,)Mx), 

where /2  (x)  is  a  quantic  of  degree  n  —  2. 

Continuing  this  process,  we  see  that  at  each  step  the 
degree  of  the  quotient  is  diminished  by  one.     Hence,  we  can 

find  n  factors  a;  — ai,  a;  — aj x  —  a^.     The  last  quotient 

will  not  involve  x,  and  is  readily  seen  to  be  cr-o,  the  coeffi- 
cient of  a;"  in /(a;). 

Now,  /W==(^-ai)/iW 

=  (a;  —  ai)(a;  —  a,)/,a; 


ao(^  —  ai)(^  —  0.2) (x  —  a,^), 


*  See  Burnside  and  Panton,  Theory  of  Equations,  2d  ed.,  Art.  195  : 
Briot  et  Bouquet,  Fonctions  Elliptiques,  Art.  23. 


GENERAL  PROPERTIES  OF  EQUATIONS.       447 

SO  that  the  equation /(a;)  =  0  may  be  written 

ao(x  —  ai)(rr  —  a^) (x  —  aj  =  0, 

which  is  evidently  satisfied  if  x  has  any  one  of  the  n  val- 
ues tti,  a^ a„. 

It  follows,  then,  that  if  every  rational  integral  equation 
has  one  root,  an  equation  of  the  nth  degree  has  n  roots. 

465.  Multiple  Boots.  The  n  roots  of  an  equation  of  the 
wth  degree  are  not  necessarily  all  different. 

Thus  the  equation  x^  —  4:X^  —  2> x  -\-  \%  =  0  may  be  writ- 
ten {x  -f  2){x  —  Z){x  —  3)  =  0,  and  its  roots  are  therefore 
-  2,  3,  8. 

The  root  3  and  the  corresponding  factor  x  —  2>  occur 
twice ;  hence  3  is  said  to  be  a  double  root.  A  root  which 
occurs  three  times  is  called  a  triple  root ;  four  times  a 
quadruple  root ;  and  so  on.  Any  root  which  occurs  more 
than  once  is  called  a  multiple  root. 

466.  Solutions  by  Trial.  When  all  the  roots  of  an  equa- 
tion but  two  can  be  found  by  trial,  the  equation  can  be 
readily  solved  by  the  process  of  §  464.  The  work  can  be 
much  abbreviated  by  employing  the  method  of  synthetic 
division  (§  462). 

Solve  the  equation 

a;*_:^-_9a;2-flla:  +  6  =  0. 
Try  +  1  and  —  1.     Substituting  these  values  for  x,  we  have 
l_l_9  +  ll+6  =  0, 
l  +  l_9-ll  +  6-0. 
Both  these  equations  are  false,  so  that  neither  +  1  nor  —  1  is  a  root. 
Try  2.     Dividing  by  a;  -  2, 

1_1_9  +  11+6|2 
+  2  +  2-14-6 

1_^1_7_    3     0 
we  see  that  2  is  a  root.     The  quotient  is  a;'  +  x^  —  7  x  —  3. 


448  ALGEBRA. 

In  this  quotient  try  2  again.     Dividing  by  «  —  2, 

1+1_7_3 

+2+6-2 


1+3_1_5 
we  see  that  2  is  not  again  a  root. 

Try  —  2.     Dividing  by  re  +  2, 

1  +  1-7-    3|-2 
-2  +  2  +  10 


1-1-5+    7 
we  see  that  —  2  is  not  a  root. 
Try  -  3.     Dividing  by  a;  +  3, 

1  +  1,7_3|_3 
-3+6+3 


1  _  2  -  1  +  0 
we  see  that  —  3  is  a  root.     The  quotient  is  x^  —  2a;  —  1. 
Hence  the  given  equation  may  be  written 

(a;-2)(a;  +  3)(a;2-2a;-l)  =  0. 

Therefore  one  of  these  three  factors  must  vanish. 
•If  a;  -  2  =  0,  a;  =  2  ;  if  a;  +  3  =  0,  a;  =  -  3  ;  if  x^  _  2a;  -  1  -  0,  we 
find  on   solving   this   quadratic  that   a;  =  1  +  \/2,    or   a;  =  1  --  V'2. 
Hence  the  four  roots  of  the  given  equation  are 

2,  -  3,  1  +  \/2,  1  -  V2. 

Exercise  133. 
Solve  the  equations : 

1.  x^ -3x^2  =  0. 

2.  a;' +  re' -16a; +  20  =  0. 

3.  a:^-8ar'^  + 21a; -18  =  0. 

4.  a;^- a:' -8a; +  12  =  0. 

5.  a;'  +  3a;'^-4  =  0. 


GENERAL    PROPERTIES    OF   EQUATIONS.  449 

6.  a;' +  2:^^-110;^ -12a: +36  =  0. 

7.  x'-x'-10x''-i-4:x-j-24:  =  0. 

8.  x'-4:x'~18x''+108x- 135  =  0. 

9.  o;^  -  40^' +160a,-2- 240:^4- 128  =  0. 

10.  o;^  -  4a;* -13a;='  + 52a;' +  36a; -144  =  0. 

11.  a;*  +  2a;^~5a;'-12n;-4  =  0. 

467.  Eoots  G-iven.     "When  all  the  roots  of  an  equation  are 
given,  the  equation  can  at  once  be  written. 

Write  the  equation  of  which  the  roots  are  1,  2,  4,  —5. 

The  equation  is         {x  -  l){x  -  2){x  -  4) (a;  +  5)  =>  0, 
or  a;^-2a;3-21a;2 +  62a;-40-0. 

468.  Relations  between  the  Eoots  and  the  OoefiScients.     The 
quadratic  equation  of  which  the  roots  are  a  and  p  is  (§  269) 

(x  -  a)(x  -  ^)  =  0 ; 
or,  multiplying  out,  , 

x^-(a  +  P)x  +  aP  =  0. 
The  cubic  equation  of  which  the  roots  are  a,  (3,  y  is 

(x-a)(x-P)(x-y)=0', 
or,      x'  -(ai-(3-{-y)x'  +  (a(3  +  ay  +  Py)x  -  a)3y  =  0. 

The  biquadratic  equation  of  which  the  roots  are  a,  /5,  y. 

Sis 

{x-a){x~Mx~y){x-h)  =  0', 
or, 

a;*-(a  +  i3  +  y  +  8)ar'  +  (a)8  +  ay  +  a8  +  )8y+^8  +  y8)a;' 

-  (a^y  +  a/38  +  ayS  +  /JyS)  X  +  a^SyS  =  0. 

And  so  on. 


450  ALGEBRA. 

Take  any  equation  in  wliich  the  highest  power  of  x  has 
the  coefficient  unity.  From  the  above  we  have  the  follow- 
ing relations  between  the  roots  and  the  coefficients  : 

The  coefficient  of  the  second  term,  with  its  sign  changed, 
is  equal  to  the  sum  of  the  roots. 

The  coefficient  of  the  third  term  is  equal  to  the  sum  of 
all  the  products  that  can  be  formed  by  taking  the  roots 
two  at  a  time. 

The  coefficient  of  ih.efou7^lh  term,  with  its  sign  changed, 
is  equal  to  the  sum  of  all  the  products  that  can  be  formed 
by  taking  the  roots  th7^ee  at  a  time. 

The  coefficient  of  the  fifth  term  is  equal  to  the  sum  of 
all  the  products  that  can  be  formed  by  taking  the  roots 
four  at  a  time  ;  and  so  on. 

If  the  number  of  roots  is  even,  the  last  term  is  equal  to 
the  product  of  all  the  roots.  If  the  number  of  roots  is 
odd,  the  last  term,  with  its  sign  changed,  is  equal  to  the 
product  of  all  the  roots. 

Observe  that  the  sign  of  the  coefficient  is  changed  when 
an  odd  number  of  roots  are  taken  to  form  a  product ;  that 
the  sign  is  unchanged  when  an  even  number  of  roots  are 
taken  to  form  a  product. 

469.  Imaginary  Eoots.  If  an  imaginary  number  is  a  root 
of  an  equation  with  real  coefficients,  the  conjugate  imagi- 
nary (§  237)  is  also  a  root. 

Let  a  +  pi,  where  i  —  V—  1,  be  a  root  of  the  equation 

the  coefficients  being  real. 

Put  a  -[-  iSi  for  X  in  the  left  member  of  the  equation,  and 
expand  the  powers  of  a  -f  fti  by  the  binomial  theorem.  All 
the  terms  which  do  not  contain  i,  and  all  the  terms  which 
contain  even  powers  of  i,  will  be  real ;  all  the  terms  which 


GENERAL  PROPERTIES  OF  EQUATIONS.       451 

contain  odd  powers  of  i  will  be  imaginary.  Representing 
the  real  part  of  the  result  by  F,  and  the  imaginary  part 
of  the  result  by  Qi,  we  have  (§  459),  since  a-\-  pi  is  a  root, 

and  therefore  P=  0  and  Q  =  0  (§  240). 

Now  put  a  —  pi  for  x  in  the  given  equation.  The  result 
may  be  obtained  from  the  former  result  by  changing  i  to 
—  i.  The  even  powers  of  i  will  be  unchanged  while  the 
odd  powers  will  have  their  signs  changed.  The  real  part 
will  therefore  be  unchanged,  and  the  imaginary  part 
changed  only  in  sign.     The  result  is 

which  vanishes,  since  by  the  preceding  P  =  0  and  Q  =  0. 

Therefore  a  —  (3i  is  a  root  of  the  given  equation  (§  459). 

This  theorem  is  generally  stated  as  follows :  Imaginary 
roots  enter  equations  in  pairs. 

It  follows  from  this  theorem  that  an  equation  of  odd 
degree  'has  always  at  least  one  real  root.  Thus  an  equa- 
tion of  the  third  degree  must  have  three  real  roots  or  one 
real  root  and  two  imaginary  roots. 

Exercise  134. 
Form  the  equations  of  which  the  roots  are 

1.  2,3,-5.  3.    2,-3,-2.  5.    3,0,-4. 

2.  3,  1,-2.  4.    3,4,-6.  6.    2,3,  i. 

7.  34-V2,  3-V2, -6.  9.    1,3,-2,-4. 

8.  1  +  V3,  1-V3,i.  10.    2,i, -2, -i. 

11.  hhl-h 

12.  I+V2,  I-V2,  V3  +  1, -V3  +  1. 

13.  2-fV=T,  2-V=n,  1  +  2V^^.  1-2V^. 

14.  1,-2,3,  -4,5.  15.   iii2.3. 


452 


ALGEBEA. 


Graphical  Representation  of  Functions. 

The  investigation  of  the  changes  in  the  value  of /(a;)  cor- 
responding to  changes  in  the  value  of  x  is  much  facilitated 
by  using  the  system  of  graphical  representation  explained 
in  the  following  sections. 


E 


470.  Oo-ordinates.  Let  X'^  and  Y'  Y  be  two  perpen- 
dicular straight  lines  drawn  in  a  plane,  intersecting  at  0. 

The  lines  X'X  and  Y'  Y 
are  called  axes  of  reference ; 
P  the   point  0  is   called   the 

I  origin. 

j  Distances  measured  from 

J ^  0  along  X'X,  as  OA,  OC, 

OE,    and    OG,   are    called 
abscissas ;     distances    meas- 
ured   from    X'X    parallel 
to  Y'Y,  as  AB,  CD,  EF, 
and  GH,  and  called  ordinates. 
Abscissas   are   considered   positive  if  measured  to   the 
right ;   negative,  if  measured  to  the  left.     Ordinates  are 
considered  positive  if  measured  upwards ;  negative,  if  meas- 
ured downwards. 

Thus,  OA,  OC,  CD,  and  EF  are  positive ;  OE,  00,  AB,  and  OB 
are  negative. 

An  abscissa  is  generally  represented  hj  x;  an  ordinate 
is  generally  represented  by  y. 

The  abscissa  and  ordinate  of  any  point  are  called  the 
co-ordinates  of  that  point.  Thus  the  co-ordinates  of  B  are 
OA  and  AB. 

The  co-ordinates  of  a  point  are  written  thus  :  {x,  y). 


GENERAL    PROPERTIES    OF   EQUATIONS.  453 

Thus,  (7,  4)  is  the  point  of  which  the  abscissa  is  7  and  the  ordi- 
nate 4. 

The  axis  JC'Xis  called  the  axis  of  abscissas,  or  the  axis 
of  X ;  the  axis  Y'  Y,  the  axis  of  ordinates,  or  the  axis  of  y. 

471.  It  is  evident  that  if  a  point  B  is  given,  its  co-ordi- 
nates referred  to  given  axes  may  be  found  by  drawing  the 
ordinate  and  measuring  the  distances  OA  and  AB. 

Conversely,  if  the  co-ordinates  of  a  point  are  given,  the 
point  may  be  readily  constructed. 

Thus,  to  construct  the  point  (7,  —  4),  a  convenient  length  is  taken 
as  a  unit  of  length.  A  distance  of  7  units  is  laid  off  on  OX  to  the 
right  from  0  to  ^.  At  -4  a  perpendicular  to  X^X  is  drawn  down- 
wards, of  length  4  units,  to  B.     Then  B  is  the  required  point. 

Construct  the  points  (3,  2) ;  (5,  4)  ;  (6,  -  3)  ;  (-  4,  -  3) ; 
(-4,2);  (-3,-5);  (4,-3). 

472.  Graph  of  a  Punotion.  Let /(a;)  be  any  function  of  x, 
where  a;  is  a  variable.  Put  y=f{x)\  then  y  is  a  new 
variable  connected  with  x  hj  the  relation  y—f(x).  If 
f{x)  is  a  rational  integral  function  of  x,  it  is  evident  that 
to  every  value  of  x  corresponds  one,  and  only  one,  value 
of  2/. 

If  different  values  of  x  be  laid  off  as  abscissas,  and  the 
corresponding  values  of  f(x)  as  ordinates,  the  points  thus 
obtained  will  all  lie  on  a  line  ;  this  line  will  generally  be 
a  curved  line,  or,  as  it  is  briefly  called,  a  curve.  This  curve 
is  called  the  graph  of  the  function /(.r)  ;  it  is  also  called  the 
locus  of  the  equation  y  =f(x). 

"We  proceed  to  construct  the  graphs  of  several  functions. 

Kemaek.  In  constructing,  or  ploiting,  as  it  is  called,  the  graph  of 
a  function,  the  student  will  find  it  convenient  to  use  the  paper  called 
plotting,  or  co-ordinate,  paper.  This  is  ruled  in  small  squares,  and 
therefore  saves  much  labor. 


454 


ALGEBRA. 


(1)   Construct  the  graph 

of3-2:r. 

Put  y  =  3  -  2a;.     The  following 

table  is  readily  computed 

If  a;=l,   y=      1. 

If  «  =  -!,   y=    5. 

"    x  =  2,   y  =  -l. 

"    re  =  -2,   y=    V. 

"    a;  =  3,   2/ =  -3. 

"    a;  =  -3,   2/=    9. 

"    re  =  4,   3/  =  — 5. 

"  -^a;  =  -4,   2/ -11. 

"    X  =  T),  y  =  -  7. 

"    .T  =  _5,   3/ =  13. 

Constructing  the  above  points,  it  appears  that  the  graph  of  the 
function  3  —  2a;  is  the  straight  line  MN. 


xr 


M 

\ 


Y 


:\ 


N 


X 


In  general,  when  the  equation  y=f{x)  contains  only  the  first 
powers  of  x  and  y,  the  locus  will  be  a  straight  line. 


GENERAL    PROPERTIES   OF   EQUATIONS. 


455 


(2)   Plot  the  graph  of  Jo;'  —  4. 

Putting  1/  =  Jx^  —  4,  we  readily  compute  the  following  table 


If 


a;=  0, 
x  =  ±l, 
x  =  ±2, 
a;  =  ±  3, 
a;  =  ±4, 
a;  =  ±5, 


-4. 
-3.5. 
-2. 
+  0.5. 

+  4. 


2/  =  +  8.5. 
y  =  +  14. 


Plotting  these  points,  we  obtain 
the  curve  here  given. 

(3)   Plot  the  graph  of 
x^  —  x^  -i-  X  —  6. 

Putting  y-=x^-x^i-x-5,  we 
compute  the  following  table.: 

If  cc  is 


0.5, 
1.0, 
1.5, 
2.0, 
2.5. 
0.0, 
-0.5, 
-1.5. 


2/ IS 

-  4.625. 

-  4.000. 

-  2.375. 
+  1.000. 
+  6.875. 

-  5.000. 

-  5.875. 
- 12.125. 


Interpolation  shows  that  if 
y  =  0,  x  =  1.88+.  Does  the  result 
agree  with  the  figure  ? 


v/ 

■ 

1 
1 

x^ 

\ 

I- 
\- 
1- 

1. 

1 
/ 

■  / 

Y^ 


473.   Consider  any  rational  integral   function   of  x,  for 
example,  x^-\-x  —  ^-}-. 

Put  3/  =  x2  +  a:  -  \^ 


456 


ALGEBRA. 


Assuming  values  of  x,  we  compute  the  corresponding 
values  of  y,  and  construct  the  graph.  Now,  any  value  of 
X  which  makes  y  =  0  satisfies  the  equation  x^-\-x  —  ^^-  =  0, 
and  is  a  root  of  that  equation ;  hence,  any  abscissa  whose 
corresponding  ordinate  is  zero  represents  a  root  of  this  equa- 
y  tion.     The  roots  may  be  found, 

approximately,  by  measuring 
the  abscissas  of  the  points  where 
the  graph  meets  XX',  for  at 
these  points  y  =  0. 

From  the  given  equation  the  fol- 
lowing table  may  be  formed : 


X^ 


X 


X  is      y  is 

Ifxis 

y  is 

0,     -  i5.75. 

-1, 

- 15.75. 

1,    -13.75. 

-2, 

-  13.75. 

2,    -   9.75. 

-3, 

-    9.75. 

3,    -    3.75. 

-4, 

-   3.75. 

4,     +    4.25. 

-5, 

+    4.25. 

The  table  shows  that  one  root  is 

between  3  and  4  (since  y  changes 

^  from  —  to  +,  and  therefore  passes 

through  zero) ;  and,  for  a  like  rea- 

/  son,  the  other  is  between  —  4  and 

-  5. 


/  474.   An  equation  of  any  de- 

gree may  be  thus  plotted,  and 
the  graph  will  be  found  to  cross 
the  axis  X'Xas  many  times  as 
■*  '  '  ^  there  are  real  roots  in  the  equa- 
tion. 

When   an   equation   has   no 
real  root,  the  graph  does  not  meet  X'X 

In  the  equation  x^  —  6x  +  13  =  0,  both  of  whose  roots  are  imagi- 
nary, the  graph,  at  its  nearest  approach,  is  4  units  distant  from  X^X. 


X'-+- 


Y' 


GENERAL    PROPERTIES    OF   EQUATIONS. 


457 


If  an  equation  has  a  dou- 
ble root,  its  graph  touches 
JT'X,  but  does  not  intersect 
it. 

The  equation  a;^  +  4  a;  +  4  =  0 
has  the  roots  —  2  and  —  2,  and 
the  graph  is  as  shown  in  the 
figure. 


X^ 


Y' 


Exercise  135. 

Construct  the  graphs  of  the  following  functions  : 
1.  a;'  +  3:r-10.     3.  ^*- 20a;' +  64.     5.  x'-bx^-\-^. 


2.  x^-2x'-\-l.      4. 


4a;+10.        6.  x^-^x^-\-x~l. 


Derivatives. 

475.  Increments.  If/(:r)  is  any  function  of  x,  then,  if 
:r  be  a  variable,  f(x)  will  also  be  a  variable.  If  we  assign 
to  X  any  particular  value  a,  f{x)  will  take  the  particular 
value /(a).  If  we  increase  x  to  a  + A,  f{x)  will  take  the 
new  value /(a  +  h).  The  increase, /(a  +  h)  —f(ct),  in  the 
value  of /(a;),  is  called  the  increment  of  f(x)  between  x  =  a 
and  x  =  a-\-  h.  In  general  the  increase, /(a; -f-  h)  —f(x),  in 
the  value  oif(x)  between  any  initial  value,  x,  of  the  varia- 
ble and  a  final  value  a;  +  A  is  called  the  increment  of  f{x) 
between  these  two  values  of  x. 

Thus,  if/(a;)  =  x^  the  increment  of /(a?)  from  x  =  2  to  a;  =  3  is 
3'  —  2^  =  27  —  8  =  19.  This  increment  is  not  equally  distributed  over 
the  interval  from  a;  =  2  to  x  =  3.  Thus,  if  we  divide  the  interval  into 
three  equal  parts  a;  =  2  to  a;  =  2J,  a;  =  2^  to  x  =  2§,  x  =  2f  to  a;  =  3, 
the  increments  of  3?  for  these  parts  are 

(2J)3  -  23  =  W.  (2 j)'  -  mf  =  -W-.  and  3^  -  {2lf  =  ^^. 

In  general,  for  equal  increments  of  x,  the  increments  off{x)  will  be 
unequal ;  that  is,  as  a;  varies,  f{x)  changes  its  values  at  a  varying  rate. 


458  ALGEBRA. 

476.   Derivatives.     The  ratio,  /(-^  +  ^)  ~.fi^)^  ^f  ^he  in- 

h 
crement  of /(:r)  to  the  corresponding  increment  of  x  is  the 
average  rate  of  change  off(x)  in  the  interval  from  xtox-j-h. 

The  limit  of  this  ratio,  as  h  approaches  0,  is  called  the 
derivative  of/(:r).  The  derivative  oif(x)  with  respect  to  x 
is  in  general  a  new  function  of  x  and  is  denoted  by  D^f(x) 
or  by/'(:r). 

"We  have  then 

Note.    A  =  0  is  read  "  as  h  approaches  0." 

The  rule  for  finding  the  derivative  of  a  function  is  there- 
fore as  follows  : 

In  the  given  function  change  x  to  x  -f-  h. 

From  the  new  value  of  the  function  subtract  the  old,  and 
divide  the  remainder  by  h. 

Take  the  limit  of  the  quotient  as  h  approaches  zero  as  a 
limit. 

The  derivative  of  a  constant  is  0,  since  the  increment  of 
a  constant  will  always  be  0. 

Find  D,(:r'). 

The  function  is  ar^. 

Change  x  to  x-{-  h,  (x -\-  hf. 

From  the  new  value  subtract  the  old,  (x  +  hy  —  x^, 

or  ?>hx''-\-2>h^x-^h\ 

Divide  by  ^.  ^x'^2>hx-\-  K\ 

Take  the  limit  as  h  approaches  0  as  a  limit,  and  we  have 

Zx\ 
:.  D,{x')^Zx\ 


GENERAL  PROPERTIES  OF  EQUATIONS.       459 

477.  Derivative  of  x".  The  function  is  x"".  Changing  x 
to  X  +  A,  we  obtain  (x  -f-  A)".  Now,  whatever  the  value  of 
n,  (x  +  hy-  can  be  expanded  by  the  binomial  theorem,  and 
we  obtain 

{x  +  hy  =  a;"  +  m-^-Vi  +  ^ii^i-lii)  :r"-^A^  + 

From  this  new  value  of  the  function  subtract  x'\  the 
old  value,  and  divide  by  h. 

We  now  have 


the  sum  of  the  terms  after  the  first  approaches  0  as  a  limit 
by  §  438. 

Hence,  to  find  the  derivative  with  respect  to  x  of  any 
power  of  X,  multiply  hy  the  exponent,  and  diminish  the 
exponent  of  x  by  one. 

Thus,  D^  (a;*)  =  4  a;^  .     jy^  ^^-^^^  =  -  3  a;"* ; 

(1)  Find  the  derivative  of  ax*". 

We  have.  D.  (ax-)  =  Jj^^^^^  p  C^ -f  ^^  -  a.t- j 

,,  limit  r(x  +  hy^-x"! 

But  this,  as  we  have  just  seen,     =^  anx^~^. 

In  general,  if  a  function  is  multiplied  by  a  constant,  its  derivative 
is  multiplied  by  the  same  constant. 

(2)  Find  the  derivative  of 


460  ALGEBRA. 

We  have 

limit  raQJx  +  h)^  +  ai{x+  A)»-»  + +  an-  apX*^ - aiX"-^ - -  anl 

^h  =  Ol  h  J 

_  limit  rgn(.v+A)»-aoa;»+ai(a;+A)**-'-  a^x*'-'^+ +«»!(■''  +  A)-«»-i"| 

limit  rao(x  +  /i)"  -  aor"1  ,  limit  faJx  -f  /i)"-i  -  aiX'^-n   , 

-h  =  ol         n        J  +  A  =  oL-^ 1 — ^J-' 

=  na^x^-'^  +  (n  —  l)aia;"-2  _^ +  ^n-i. 

That  is,  to  differentiate  any  rational  integral  function,  we  differ- 
entiate each  term  separately,  and  add  the  results  together. 


Exercise  136. 

Find  the  derivatives  of  the  following  functions  : 

1.    x\       2.    x\        3.    x\       4.    dx\       5.    x-\       6.    -• 

x^ 

7.  x^^2x.     8.  x^  +  3a;  +  4.     9.   3.i^*4-2;r^  +  5:c'  +  6:r+4. 

1 


10.    (a; +  3)1     11.    {x^a)\     12.    (:r+l)-l     13. 


a;^-l 


478.  Derivative  of  a  Product.     Suppose  that  the  product 
consists  of  three  factors, 

f{x)  =  {x  —  a){x  —  b){x  —  c). 

We  have 

DJ{x) 

=  limit  nx+h-a)(x+k-b){x+h-c)-{x-a){x~b){x-c)'\ 


GENEBAL  PROPERTIES  OF  EQUATIONS.       461 

or,  if  we  denote  x  —  a  hy  a,  x  —  b  hy  P,  x  —  c  hy  y, 
DJ(x) 

_  limit  r(a+A)(y8+A)(Y  +  A)-affy"| 

^=^L  ^  J 

^  limit  rafty  +  h(a/3  +  py-{-ya)-i-h'(a  +  P  +  y)  +  h'-afiyl 

^ limit  rh(a(3  +  /3y  +  7«^)  +  ^'(^  +  )8  +  y)  +  AH 
^=^L  ^  J 

-a/8  +  y8y  +  ya 

—  (x—  a)(x  —  b)-\-  (x  —  b){x  —  c)  -\-  {x  —  c)(x  —  a). 
Similarly  if  f(x)  consist  of  four  factors, 
f  (x)  =  (x  — a)(x  —  b)(x  —  c)(x  —  d), 

f  (x)  =  DJ{x)  =  (x-b){x-  cXx  -d)  +  (x-cXx-  d){x  -  o) 
-\-{x  —  d)(x  —  a)(x  —  b)-}-(x  —  a){x  —  b){x  —  c), 
and  in  general  we  have  the  result : 

The  derivative  of  any  product  of  n  factors 

(x  —  a){x  —  b){x  —  c){x  —  d)  

is  the  su7n  of  n  terms,  each  of  which  consists  of  the  product 
of  n  —  1  of  the  same  factors. 

479.   Multiple  Factors.     Suppose  that 

f{x)  =  (x  —  a){x  —  b){x  —  c) 

as  before,  but  let  b  =  a.     We  have  then  from  the  preced- 
ing section 

fx  =  DJ{x)  =  {x-  a){x  -  b)^{x-  b){x  -c)-\-(x-  c)(x-a) 
=  (x  —  ay  -{-(x  —  a)(x  ~c)  +  (x  —  c)(x  —  a), 
and  a;  —  (2  is  a  factor  of  f'(x)  as  well  as  of  f(x). 


462  ALGEBRA. 

Similarly  in  the  case  of  four  factors, 

f{x)  =  (x  —  a){x  —  b){x  —  c){x  —  d), 

every  term  of  f'(x)  contains  either  x  —  aorx  — bsiSSi  fac- 
tor. Consequently  when  b  =  a,  f\x)  will  contain  a;  —  a  as 
a  factor  in  every  term. 

In  general,  we  have  the  proposition 

If  f(x)  contain  a  double  factor  (x  —  a)'\  f'(x)  ivill  con- 
tain the  same  factor  as  a  simple  factor. 

By  the  same  reasoning  it  is  shown  that  if  f(x)  contain  a 
triple  factor  (x  —  a)^,  f'(x)  will  contain  (x  —  ay  as  a  factor, 
and  in  general 

If  f  (x)  contain  a  factor  (x  —  a)*,  L'(x)  will  contain  as  a 
factor  (x  —  a)*~\ 


480,  Multiple  Eoots.  Let  the  equation  f(x)  =  0  have  a 
multiple  root  a  of  order  k.  Then  f{x)  contains  (x  —  a)*  as 
a  factor,  and  consequently  f'{x)  contains  (x  —  a)"'^  as  a 
factor.  The  H.  0.  F.  of  f(x)  and  f'(x)  therefore  contains 
(x  —  ay~^  as  a  factor. 

To  find  the  multiple  roots  of  the  equation  f(x)  —  0. 

Find  the  H.  0.  F.  oif(x)  and  f{x)  and  resolve  it  into  its 
factors.  Each  multiple  root  will  occur  once  more  in  the 
equation  f{x)  =  0  than  the  corresponding  factor  occurs  in 
the  H.  0.  F. 

Find  the  multiple  roots  of  the  equation 

a^  -  Ibx'  -  10a;'  +  60x  -f  72  =  0. 

Here  f  {x)  =  a^ -I5a^ -lOx^ +  60x +  12, 

f{x)  =  5  x*  -  45  x2  -  20  X  +  60. 


GENERAL    PROPERTIES   OF   EQUATIONS. 


463 


Fiad  the  H.  C.  F.  of/(x)  aud/(a;),  as  follows : 


5)5  +  0-45-20  +  60 
1  +  0-   9-   4  +  12 

1  +  0-15-10  +  60  +  72 
1  +  0-   9-   4  +  12 

1+1_   8-12 

_6)-   6-   6  +  48  +  72 

-1-    1+    8  +  12 
_1_    1+    8  +  12 

1+    1-   8-12 

1-1 


.-.  The  H.  C.  F.  is  ar'  +  a;2  -  8a;  -  12. 

We  find,  by  trial,  that  3  is  a  root  of  the  equation 

x3  +  x2-8a;-12  =  0. 

The  other  roots  are  found  to  be  —  2  and  —  2. 

Hence  t^  +  x"  -  8x -12  =  {x -Z)(x  +  2f. 

Therefore  3  is  a  double  root  and  —  2  a  triple  root  of  the  given 
equation.  As  the  equation  is  of  the  fifth  degree,  these  are  all  the 
roots,  and  the  equation  may  be  written 

{x  -  3y{x  +  2)3  =.  0. 

Having  found  the  multiple  roots  of  an  equation,  we  may 
divide  by  the  corresponding  factors,  and  find  the  remaining 
roots,  if  any,  from  the  reduced  equation. 


Exercise  137. 

The  following  equations  have  multiple  roots.     Find  all 
the  roots  of  each  equation. 

1.  a^-bx'  +  lx-^^O. 

2.  0:^-3x^  +  4  =  0. 

3.  x'-2x^-1x''-j-20x-12  =  0. 

4.  x'-2x'~Ux'+12x-\-S6  =  0. 

5.  a;*-24a;^  +  64a;-48  =  0. 

6.  a:^  +  n'*-17a;'-21a;^  +  72a:+108  =  0. 

7.  a;« -5a:^  +  5x*  +  9a;'-14a;'-4a:-f  8  =  0. 


464  '  ALGEBRA. 


Transformation  of  Equations. 

481.  The  solution  of  an  equation,  and  the  investigation 
of  its  properties,  is  often  facilitated  by  a  change  in  the  form 
of  the  equation.  Such  a  change  of  form  is  called  a  trans- 
formation of  the  equation. 

482.  Eoots  with  Signs  changed.  The  roots  of  the  equation 
f(— x)  =  0  are  those  of  the  equation  f(x)  =  0,  each  with 
its  sign  changed. 

For,  let  a  be  any  root  of  equation  f(x)  —  0. 

Then,  we  must  have  /(a)  =  0. 

In  the  quantic/(—  x)  put  —  a  for  r^ ;  that  is,  a  for  —  x. 

The  result  is /(a). 

But  we  have  just  seen  that  /(a)  vanishes,  since  a  is  a 
root  of  the  equation  f(x)  =  0.  Hence,  /(—  x)  vanishes 
when  we  put  —  a  for  x,  and  (§  459)  —  a  is  therefore  a  root 
of  the  equation  /(—  x)  =  0. 

To  obtain  thef(—x)  we  change  the  sign  of  all  the  odd 
powers  of  x  in  the  quantic/(a;). 

Thus,  the  roots  of  the  equation 

a;*-2a;3-13a;2  +  14x  +  24  =  0 
are  2,  4,  —  1,  —  3  ;  and  those  of  the  equation 

X*  +  2a;3  -  13a;2  -  14a;  +  24  =  0 
are  -  2,  -  4,  +  1,  +  3. 

483.  Eoots  multiplied  by  a  Given  Number.  Consider  the 
equation 

ax*  +  bx'  -\-cx'-\-dx-\-e  =  0.  (1) 

Put  y  —  mx,  then  x  —  —  ;  and  the  equation  becomes 


GENERAL  PROPERTIES  OF  EQUATIONS.       465 

Let  a  be  any  root  of  (1)  ;  the  left-member  of  (1)  vanishes 
when  we  put  a  for  x,  and  we  obtain 

aa*  +  ha!'  +  ca^  +  C?a  +  e  =  0. 

In  the  left-member  of  (2),  put  ma  for  y  ;  we  obtain 

which,  as  we  have  just  seen,  vanishes.     Hence,  if  a  is  a  root 
of  (1),  ma  is  a  root  of  (2). 

Similarly,  for  an  equation  of  any  degree. 

Equation  (2)  may  be  written  in  the  form 

ay*  _|_  'mby^  -f-  jrfcy'^  +  m!dy  -f-  tn^e  =  0. 

The  above  form,  if  written  with  x  in  place  of  y,  gives 
the  following  rule : 

Multiply  the  second  term  by  m ;  the  third  term  by  m^ ; 
and  so  on.  Zero  coefficients  are  to  be  supplied  for  missing 
powers  of  x. 

Write  the  equation  of  which  the  roots  are  the  doubles  of 
the  roots  of  the  equation 

3x*  -2x'-\-4:x'-e>x-5  =  0. 

Here  m  =  2,  and  the  result  is 

3x*  -  2(2)x3  +  4(2)2a;2  -6{2fx  -  5(2)*  =  0, 
or  3a;*-4a'3  +  16a;2-48a;-80  =  0. 

484.  Removal  of  Fractional  Coefficients.  If  any  of  the  co- 
efficients of  an  equation  in  the  form 

X''  +jOia:"-^  -{-p,x''-'  -f pn  =  0 

are  fractions,  we  can  remove  fractions  as  follows : 

Multiply  the  roots  by  m ;  then  take  m  so  that  all  of  the 
coefficients  will  be  integers. 

Note.  Every  equation  can  be  reduced  to  this  form,  called  thep 
form,  by  dividing  through  by  the  coeflBcient  of  a;**. 


466  ALGEBRA. 

Reduce  to  an  equation,  in  the  p  form,  with  integral  coefficients, 

Dividing  by  2,  a;^  -  ^  a;^  +  ^^^  x  +  |  -=  0. 

Multiplying  the  roots  by  m  (^  483), 

6  12  8 

The  least  value  of  m  that  will  render  the  coefficients  all  integral 
is  seen  to  be  6.     Putting  6  for  m,  we  obtain 

a;=^-a;2^  15a; +  27  =  0, 

the  equation  required. 

Any  multiple  of  6  might  have  been  used  instead  of  6,  but  the 
smaller  the  number,  the  easier  the  work. 

485.    Reciprocal  Equations.     Consider  the  equations 

nx*  +  bx'  -j-cx'-\-dx-\-e  =  0,  (1) 

ex*  +  dx'  +  cx^  +  bx  +  a  =-  0.  (2) 

The  coefficients  in  the  one  equation  are  the  same  as  those 
in  the  other  reversed  in  order. 

If  a  be  any  root  of  (1),  then  -  will  be  a  root  of  (2).  For 
if  a  be  a  root  of  (1),  we  have 

aa*  +  ba'  +  ca'  +  da  +  e  =  0  ;  (3) 

and  if  -  be  a  root  of  (2),  we  have 
a 

e    ,   d   ,    c    ,   b   ,  r\ 

a        a'       a        a 

or  e-\-dai-  ca"  +  ba'  +  aa*  =  0.  (4) 

But  the  equations  (3)  and  (4)  are  identically  the  same. 

The  four  roots  of  equation  (2)  are  therefore  the  recip- 
rocals of  the  four  roots  of  equation  (1). 

Similarly  for  equations  of  any  degree. 

The  coefficients  of  an  equation  may  be  such  that  revers- 
ing their  order  does  not  change  the  equation.     In  this  case 


GENERAL  PROPERTIES  OF  EQUATIONS.       467 

the  reciprocal  of  a  root  is  another  root  of  the  equation. 
That  is,  one  half  the  roots  are  reciprocals  of  the  other  half. 
An  equation  in  which  this  is  true  is  called  a  reciprocal 
equation. 

(1)  Write  the  equation  of  which  the  roots  are  the  recip- 
rocals of  the  roots  of 

2a;5  -j-  4x*  -  3x'  -^  7  f  -\-  2.x  -  5  =  0. 

The  result  is 

-  5:^^  +  2x'  +  7n-'  -  3a;2  +  4n'  +  2  =  0, 
or  5a^-  2x*  -  7x'  +  Sx'  -  4a;  -  2  =  0. 

(2)  The  equation  Qx' -29x'  +  27 x" +  27 x'-29x+6  =  0 
in  a  reciprocal  equation. 

Its  roots  are  found  to  be  —  1,  2,  3,  ^,  i.  Here  —  1  is 
the  reciprocal  of  itself;  J  is  the  reciprocal  of  2 ;  and  I  of  3. 

486.  Boots  diminished  by  a  Given  Number.  Consider  the 
equation 

ax'  +  bx'  -{-cx''-i-dx  +  e=^0.  (1) 

To  obtain  the  equation  which  has  for  its  roots  the  roots 
of  the  above  equation  each  diminished  by  h,  we  proceed 
as  follows  : 

Put  y  =  X  —  h;  then  x  =  y  -\-  h;  and  the  equation  be- 
comes 

a{y  +  hy  +  h{y  +  Kf-\-c{y  +  Kf  +  d{y  +  h)  +  e  =  0.    (2) 

Let  a  be  any  root  of  (1)  ;  then  we  must  have 

aa'  +  ha"  +  co?-\-da-\-e  =  0. 

In  the  left-member  of  (2)  put  a  for  y  -f  ^ ;  that  is, 
a  —  A  for  3/ ;  we  obtain 

aa*  +  ^a'  -\-  Co?  -\-  da -\-  e ; 

which,  as  we  have  just  seen,  vanishes. 


468  ALGEBRA. 

Hence,  a  —  A  is  a  root  of  (2).  Since  the  above  is  true  for 
each  of  the  roots  of  (1),  and  the  two  equations  are  evidently 
of  the  same  degree,  the  roots  thus  found  are  all  the  roots 
of  equation  (2). 

Similarly  for  an  equation  of  any  degree. 

487.  Calculation  of  the  Coefficients.  The  labor  of  calculat- 
ing the  coefficients  in  the  transformed  equation  of  (§  486) 
can  be  greatly  reduced  by  Synthetic  Division. 

Suppose  equation  (2)  of  (§  486)  to  be  expanded  and 
arranged  in  order  of  the  powers  of  y,  and  suppose  the  result 

Ay'  -\-  Bf  -\-  Cy'  -\-  Dy  ^  E=^. 
Since  y  =  x—h,  this  equation  is  equivalent  to 
A{x-hy-{-B{x--Kf-\-C(ix-Kf-{'D{x-h)+E=0.  (3) 

If  we  divide  the  left-hand  member  of  (3)  by  a;  —  A,  the 
remainder  is  E,  the  last  coefficient.  And  if  we  divide 
the  resulting  quotient  again  by  x  —  h,  the  remainder  is  D, 
the  next  to  the  last  coefficient,  and  so  on. 

Now  the  left-hand  member  of  (3)  is  identically  the  same 
as  the  left-hand  member  of  (1).  Consequently  we  have  the 
following  rule  for  transforming  any  equation  f(x)  =  0  into 
an  equation  whose  roots  are  less  by  h. 

Divide /(:c)  by  a:  —  A,  and  the  remainder  will  be  the  last 
coefficient  in  the  transformed  equation.  Divide  again  by 
x  —  h,  and  the  remainder  will  be  the  coefficient  of  the  last 
term  but  one  in  the  transformed  equation.  Continue  the 
process  until  all  the  coefficients  are  determined. 

Obtain  the  equation  which  has  for  its  roots  the  roots  of 
the  equation 

3^4  -  7a;'  ~  4:r'^  -  6ar  +  5  =  0, 

each  diminished  by  3. 


GENERAL  PROPERTIES  OF  EQUATIONS.       469 


The  work  is  as  follows  : 


3_    7_   4_     6  +  5[3^ 
+    9+    6+      6+0 

3+2+2+      0  +  5         (5  =  first  remainder.)    ^ 
+    9  +  33  +  105 

3  +  11  +  35  +  105  (105  =  second  remainder.) 

9  +  60 


3  +  20  +  95  (95  =  third  remainder.) 


3  +  29  (29  =  fourth  remainder.) 

The  required  equation  is  then 

3a:*  +  29a:3  +  95a;2  +  105a;  +  5  =  0. 

Exercise  138. 

Multiply  the  roots  of  each  of  the  following  equations  by 
the  number  placed  opposite  the  equation  : 

1.  x'-5x''-i-2x  —  S  =  0  (-1). 

2.  x*-{-4:x''-{-3x  +  b  =  0  (+2). 

3.  2x3-3a;'^  +  5:r-7-0  (-2). 

4.  5x*-3x'-{-2x-Q  =  0  (+5). 

Write  the  equations  which  have  for  their  roots  the  recip- 
rocals of  the  roots  of  the  following  equations  : 

5.  2x'-\-3x''-x-2  =  0. 

6.  Sx*-i-5a^-x^  +  2x  +  3  =  0. 

7.  2s^i-Sx'-Qx^i-6a^-3x-2  =  0. 

Diminish  the  roots  of  each  of  the  following  equations  by 
the  number  opposite  the  equation  : 

8.  x'  +  ^x'-Ux-n^^O  (+2). 

9.  Sx*-10x'  +  2x'-bx-\-6^0         (+3). 
10.    x'-bx^-^x'-\-Sx-\-10  =  0  (-2). 


470  ALGEBRA. 

488.  Descartes'  Eule  of  Signs.  An  equation  in  which  all 
the  powers  of  x  from  x^  to  a;"  are  present  is  said  to  be  com- 
plete ;  if  any  powers  of  x  are  missing,  the  equation  is  said 
to  be  incomplete.  An  incomplete  equation  can  be  made 
complete  by  writing  the  missing  powers  of  x  with  zero 
coefficients. 

A  permanence  of  sign  occurs  when  -\-  follows  -f,  or  — 
follows  -- ;  a  variation  of  sign  when  —  follows  +,  or  -f  fol- 
lows — . 

Thus,  in  the  complete  equation 

x«  -  3 a;5  +  2a;*  +  a;3  -  2x2  -  a;  -  3, 
writing  only  the  signs 

+     -     +     +     ---, 
we  see  that  there  are  three  variations  of  sign  and  three  permanences. 
For  positive  roots,  Descartes'  rule  is  as  follows : 

The  number  of  positive  roots  of  the  equation  f  (x)  =  0  can- 
not exceed  the  number  of  variations  of  sign  in  the  quantic 
f(z). 

To  prove  this  it  is  only  necessary  to  prove  that  for  every 
positive  root  introduced  into  an  equation  there  is  one  varia- 
tion of  sign  added. 

Suppose  the  signs  of  a  quantic  to  be 

+     -     +     +     + + 

and  introduce  a  new  positive  root.    We  multiply  by  x—h; 
or,  writing  only  the  signs,  by  -| .     The  result  is 

+     -     +     +     +     --     + 
+     - 

+     -     +     +     +     --     + 
-     +     ---     +     +     - 

4--    +    ±±    —    ^=    +    — 


GENERAL    PROPERTIES    OF    EQUATIONS.  471 

The  ambiguous  signs  zt,  =F  indicate  that  there  is  doubt 
whether  the  term  is  positive  or  negative.  Examining  the 
product  we  see  that  to  permanences  in  the  multiplicand 
correspond  ambiguities  in  the  product.  Hence,  we  cannot 
have  a  greater  number  of  permanences  in  the  product  than 
in  the  multiplicand,  and  may  have  a  less  number.  But 
there  is  one  more  term  in  the  product  than  in  the  multipli- 
cand. Hence  we  have  at  least  one  more  variation  in  the 
product  than  in  the  multiplicand. 

For  each  positive  root  introduced  we  have  at  least  one 
more  variation  of  sign.  Hence  the  number  of  positive  roots 
cannot  exceed  the  number  of  variations  of  sign. 

Negative  Boots,  Change  x  to  —  x.  The  negative  roots 
of  the  given  equation  will  be  positive  roots  of  this  latter 
equation  (§  482),  and  the  preceding  rule  may  then  be 
applied. 

489.  From  Descartes'  rule  we  obtain  the  following : 

If  the  signs  of  the  terms  of  an  equation  are  all  positive, 
the  equation  has  no  positive  root. 

If  the  signs  of  the  terms  of  a  complete  equation  are  alter- 
nately positive  and  negative,  the  equation  has  no  negative 
root. 

If  the  roots  of  a  complete  equation  are  all  real,  the  num- 
ber of  positive  roots  is  the  same  as  the  number  of  varia- 
tions of  sign,  and  the  number  of  negative  roots  is  the  same 
as  the  number  of  permanences  of  sign. 

490.  Existence  of  Imaginary  Roots.  In  an  incomplete 
equation  Descartes'  rule  sometimes  enables  us  to  detect  the 
presence  of  imaginary  roots. 

Thus,  the  equation  a;^  +  5a;  +  7  =  0 

may  be  written  3?  ±  Ox^  +  5a;  +  7  =  0. 


472  ALGEBRA. 

We  are  at  liberty  to  assume  that  the  second  term  is  positive,  or 
that  it  is  negative. 

Taking  it  positive,  we  have  the  signs 

+         +         +         +  ; 

there  is  no  variation,  and  the  equation  has  no  positive  root. 
Taking  it  negative,  we  have  the  signs 

+         -         +         +  ; 

there  is  but  one  permanence,  and  therefore  not  more  than  one  nega- 
tive root. 

As  there  are  three  roots,  and  as  imaginary  roots  enter  in  pairs,  the 
given  equation  has  one  real  negative  root  and  two  imaginary  roots. 

Exercise  139. 

All  the  roots  of  the  equations  given   below  are   real ; 
determine  their  signs. 

1.  a;H  40;"- 43a;' -58:r  + 240  =  0. 

2.  a;'-22:r'  +  155rc-350-0. 

3.  a:*  +  4a;' -35a;' -78a; +  360  =  0. 

4.  a;' -12a;' -43a; -30  =  0. 

5.  a;^-3a;*-5a;'  +  15a;'  +  4a;-12  =  0. 

6.  a;3  -  12a;' +  47a; -60  =  0. 

7.  Show  that  a;®— 3a;'  —  a;-|-l  =  0  has  at  least  two  im- 

aginary roots. 

8.  Show  that  a;*  +  15a;'-|-7a;  —  11  =  0  has  two  imaginary 

roots,  and  determine  the  signs  of  the  real  roots. 

9.  Show  that  a;**  —  1  =  0  has  but  two  real  roots,  +  1  and 

—  1,  when  n  is  even ;  and  but  one  real  root,  +  1, 
when  n  is  odd. 

10.    Show   that  a;"  + 1  =  0  has  no  real  root  when  n  is 
even  ;  and  but  one  real  root,  —  1,  when  n  is  odd. 


CHAPTER  XXXVI. 
NUMERICAL   EQUATIONS. 

491.  A  real  root  of  a  numerical  equation  is  either  com- 
mensurable or  incommensurable. 

Commensurable  roots  are  either  integers  or  fractions. 
Repeating  decimals  can  be  expressed  as  fractions,  and  roots 
in  that  form  are  consequently  commensurable. 

Incommensurable  roots  cannot  be  found  exactly,  but 
may  be  calculated  to  any  desired  degree  of  accuracy  by  the 
method  of  approximation  explained  in  this  chapter. 

Commensurable   Roots. 

492.  Integral  Boots.  The  process  of  finding  integral  roots 
given  in  §  466  is  long  and  tedious  when  there  are  many 
numbers  to  be  tried.  The  number  of  divisors  to  be  tried 
is  diminished  by  the  following  theorem  : 

Every  integral  root  of  an  equation  ivith  integral  coefficients 
is  a  divisor  of  the  last  term. 

We  shall  prove  this  for  an  equation  of  the  fourth  degree, 
but  the  proof  is  perfectly  general. 

Let  h  be  an  integral  root  of  the  equation 

ao;*  +  hx^  -[- cx^  -\-  dx -{-  e^O, 

where  the  coefficients  a,  b,  c,  d,  e  are  all  integers. 
Since  A  is  a  root, 

ah*  +  bh'  +  ch'-}-dh-\-e.=  0,  (§  459) 

or,  e  =  —  dh  —  ch^  —  bh^  —  ah*. 


'4 

ALGEBRA, 

Dividing  by  h, 

e  _ 

-d-ch-bh' 

ah' 


Since  the  right  member  is  an  integer,  the  left  member 
must  be  an  integer.     That  is,  e  is  divisible  by  h. 

Similarly,  for  any  equation  with  integral  coefficients. 

Hence,  in  applying  the  method  of  §  466,  we  need  try 
only  divisors  of  the  last  term.  The  necessary  labor  may 
be  still  further  reduced  by  the  method  shown  in  the  fol- 
lowing section. 

Find  the  integral  roots  of  the  equation 

2x'  -  a;'  -  29:r'^  +  34:c  +  24  -  0. 

Neither  +  1  nor  —  1  is  a  root. 

The  other  divisors  of  24  are  ±  2,  ±  3,  ±  4,  ±  6,  ±  8,  ±  12,  ±  24. 

Try +2:  2 -  1  -  29  +  34  +  24[2_ 

+  4+6-46-24 


2  +  3-23-12+    0 


Hence  +  2  is  a  root.    Dividing  the  equation  by  a;  —  2,  we  have  for 
the  resulting  equation 

2a;3  +  3a;2_  23a; -12-0. 

Try +3:  2  +  3-23-12|^ 

6  +  27  +  12 

2+9+4+0 
Hence  3  is  a  root. 
The  remaining  roots  are  roots  of  the  quadratic  equation 

2a;2  +  9a;  +  4  =  0, 
and  are  —  4  and  —  ^. 

Therefore  the  roots  of  the  given  equation  are  2,  3,  —  4, 

~2' 


NUMERICAL    EQUATIONS.  475 

493.  Fractional  Koots.  An  equation  with  iyitegral  coeffi- 
cients, in  which  the  coefficient  of  the  highest  power  of  x  is 
unity,  cannot  have  a  rational  fraction  for  a  root. 

The  general  form  of  such  an  equation  is 

a;"  -\-p,x^-'  -{-p.x^-'  + +^„  -  0, 

where  pi,  p^, />„  are  integers. 

If  possible  let  -,  where  h  and  k  are  integers,  and  -  is  in 
its  lowest  terms,  be  a  root.     Then, 

Jn+P^^.+P-^-j^.+ Pn-0' 

Multiplying  by  k""'^  and  transposing, 

^=-p,h^-'  -p,Jt-'k  -  • pJc'^-K 

Ic 

Now  the  right  member  is  an  integer ;  the  left  member  is 
a  fraction  in  its  lowest  terms,  since  A"  and  ^  have  no  common 
divisor  as  h  and  k  have  no  common  divisor,  §  415,  V.  But 
a  fraction  in  its  lowest  terms  cannot  be  equal  to  an  integer. 
Hence  -,  or  any  other  rational  fraction,  cannot  be  a  root. 

The  real  roots  of  an  equation  with  integral  coefficients  in 
the  p  form  are,  therefore,  integral  or  incommensurable. 

In  case  an  equation  has  fractional  roots,  we  can  find  them 
as  follows  : 

Transform  the  equation  into  an  equation  with  integral 
coefficients  by  multiplying  the  roots  by  some  number  m 
(§  484).  Find  the  integral  roots  of  the  transformed  equa- 
tion, and  divide  each  by  ni. 

Solve  the  equation 

36a;*-55a;^-35^-6  =  0. 


476  ALGEBRA. 

Write  this  x*  -  f  |  x"  -  -|f  a;  -  ^  =  0. 

Multiplying  the  roots  by  6,  we  obtain 

a;* -55x2 -210  a; -216  =  0, 
of  which  the  roots  are  found  to  be  —  2,  —  3,  —  4,  9. 
Hence,  the  roots  of  the  given  equation  are 

2  3  49.    or  1  1  23 

Exercise  140. 

Find  the  commensurable  roots  and,  if  possible,  all  the 
roots  of  the  following  equations  : 

1.  x^-^r^x'-AQx-^^^-^Q. 

2.  a;'-3:r'-10a;  +  24  =  0. 

3.  x'-\-x'^-^^x''-2Ax-12  =  0. 

4.  a;* -7a;''- 6:^^-18:^+16-0. 

5.  x'-^a^-\-Vlx''^21x-m  =  0. 

6.  18:r*-33:c^-13a;^  +  12a;  +  4  =  0. 

7.  27a;*-72a;^  +  33a;'  +  8a;-4  =  0. 

8.  36:r^  -  60:u*  -  167  a;^-f52:r'  +  57:r- 18  =  0. 

Incommensurable  Roots. 

494.  Location  of  the  Eoots.  In  order  to  calculate  the 
value  of  an  incommensurable  root  we  must  first  find  a 
rough  approximation  to  the  value  of  the  root ;  for  example, 
two  integers  between  which  it  lies.  This  can  generally  be 
accomplished  by  the  aid  of  the  following 

Theorem  on  Change  of  Sign.  Let  two  real  numbers  a  and 
b  he  put  for  x  in  f  (x).  If  the  resulting  values  of  f  (x)  have 
contrary  signs,  an  odd  number  of  roots  of  the  equation 
f  (x)  —  0  lie  between  a  aiid  b, 


NUMERICAL    EQUATIONS. 


477 


As  X  changes  from  a  to  h,  passing  through  all  interme- 
diate values,  f{x)  will  change  from  /(a)  to  f{h),  passing 
through  all  intermediate  values.  Now,  in  changing  from 
f{a)  to  f(b),f(x)  changes  sign. 

Hence, /(a:)  must  pass  through  the  value  zero.  That  is, 
there  is  some  value  oi  x  between  a  and  b  which  causes /(a:) 
to  vanish ;  that  is,  some  root  of  the  equation  f(x)  —  0  lies 
between  a  and  b. 

But  f(x)  may  pass  through  zero  more  than  once.  To 
change  sign, /(a;)  must  pass  through  zero  an  odd  number  of 
times;  and  an  odd  number  of  roots  must  lie  between  a 
and  b. 

Applied  to  the  graph  of  the  equation,  since  to  a  root  cor- 
responds a  point  in  which  the  graph  meets  the  axis  of  x 
(§  473),  the  above  simply  means  that  to  pass  from  a  point 
below  the  axis  of  x  to  a  point  above  that  axis,  we  must  cross 
the  axis  an  odd  number  of  times. 

In  some  examples  Descartes'  Rule  of  Signs  may  be  of  use. 


Example.     The  equation 

x'  -  2x'  -  llo;^  +  6x  +  2  --=  0 

has,  by  Descartes'  rule  (§  488),  not  more  than  two  positive 
roots  and  not  more  than  two  negative  roots. 


Wefind(H63),    /(O)  =  +    2 

m=-  4 

/(2)  =  -30 
/(3)  =  -52 
/(4)  =  -22 


/(5)  =  +  132 
/(-1)  =  -  12 
/(-2)  =  -  22 
/(-3)  =  +  20 
/(-4)=  +  186, 


Hence  there  are  two  positive  roots,  one  between  0  and  1,  and  one 
between  4  and  5 ;  and  two  negative  roots,  one  between  0  and  —  1, 
and  one  between  —  2  and  —  3. 


478  ALGEBRA. 

Let  us  find  more  closely  a  value  for  the  root  between  0  and  1. 
We  find  /(0.5)  =  +  2.06+.  Since  /(I)  =  -  4,  the  root  lies  between 
0.5  and  1. 

Try  0.8  :  we  find  /(0.8)  =  -  0.9-f .  Hence  the  root  lies  between 
0.5  and  0.8. 

We  find /(0. 7)  =  +  0.4+.    Hence  the  root  lies  between  0.7  and  0.8. 

In  a  similar  manner  we  find  the  root  between  0  and  —  1  to  lie 
between  —  0.2  and  —  0.3. 

The  first  significant  figures  of  the  roots  are  accordingly  0.7,  4, 
-  0.2,  -  2. 

Exercise  141. 

Determine  the  first  significant  figure  of  each  real  root  of 
the  following  equations : 

1.  x''-x'-2x-{-l  =  0.  3.   x'-bx'-{-7-=0. 

2.  x'  —  5x-B-=-0.  4.    x'~\~2x''-S0x-{-S9  =  0. 

5.  x'  —  6x'-{-Sx-}-5=-0. 

6.  ^^-^  _|_  9^2  _^  24a; +17  =  0. 

7.  a;' -  15a;' +  63a; -50  =  0. 

8.  a;*-8a;'  +  14a;''  +  4a;-8  =  0. 

495.  Horner's  Method,  Positive  Roots.  Suppose  the  first 
figure  of  the  root  to  have  been  found.  Any  number  of 
remaining  figures  may  be  calculated  by  the  method  of 
approximation  known  as  Horner's  Method. 

We  proceed  to  illustrate  the  process  by  an  example. 

Take  the  equation 

x'-6x'  +  Sx-{-6  =  0.  (1) 

By  §  494  one  root  of  this  equation  lies  between  1  and  2. 
We  proceed  to  calculate  that  root. 


NUMERICAL    EQUATIONS.  479 

Diminish  the  roots  by  1  (§  486) : 

1         -6         +3         +5    IJ, 
+1-5-2 

—5        -2        +3 

±1      zl 

-4        -6 

+  1 
—  3 

The  transformed  equation  is,  therefore, 

y'-3?/-6y  +  3  =  0.  (2) 

The  roots  of  equation  (2)  are  each  less  by  1  than  the 
roots  of  equation  (1).  Equation  (1)  has  a  root  between  1 
and  2 ;  equation  (2)  has,  therefore,  a  root  between  0  and  1. 
Since  this  root  is  less  than  1,  ^/^  and  y^  are  both  less  than  y. 
Neglecting  these  terms,  we  have 

_6y  4-3  =  0,  ory  =  0.5. 

At  this  stage  of  the  process  the  figure  thus  obtained  will 
not  in  general  be  the  correct  one.  If,  however,  we  neglect 
only  the  y^  term,  we  obtain 

-3y^-6y+3  =  0, 

or  /  +  2y-l-0, 

of  which  one  root  is  V2  —  1  =  0.4+. 

We  can  also  find  the  second  figure  of  the  root  as  foUowa : 

Take  the  first  value  0.5. 

With  this  assumed  value  of  y,  computing  the  value  of  y'^-Sy^ 
and  substituting,  we  obtain  63/ =  2.375;  whence  y  =  0.4,  approxi- 
mately. 


480  ALGEBRA. 

We  now  diminish  the  roots  of  (2)  by  0.4  : 

1         -3  -6  +3        104 

+  0.4  -1.04         -  2.816 

\       —2.6  -7.04         +0.184 

\      +0.4         -0.88 
'      —2.2  -  7.92 

+  0.4 
—  1.8 
The  second  transformed  equation  is 

2^-1.82^  -  7.920  +  0.184  =  0.  (3) 

The  roots  of  (3)  are  less  by  0.4  than  those  of  (2),  and  less 

by  1.4  than  those  of  (1).     Equation  (2)  has  a  root  between 

0.4  and  0,5  ;  equation  (3)  has,  therefore,  a  root  between  0 

and  0.1. 

Since  this  root  is  much  less  than  1,  we  shall  probably 
obtain  a  correct  value  for  the  next  figure  of  the  root  by 
neglecting  the  ^  and  ^  terms  in  equation  (3). 

This  gives  -7.922  +  0.184  =  0;  whence  2  =  0.02+. 
Diminish  the  roots  of  (3)  by  0.02  : 

1        -1.8  -7.92  +0.184       10.02 

+  0.02        -  0.0356        -0.159112 

-  1.78         -  7.9556         +  0.024888 

+  0.02    ■    -  0.0352 

+  1.76         -7.9908 

+  0.02 

-1.74 
The  third  transformed  equation  is 

?/  -  1.74 u"^  ~  7.9908 u  +  0.024888  =  0.  (4) 

The  roots  of  (4)  are  less  by  0.02  than  those  of  (3),  and 
less  by  1.42  than  those  of  (1). 

Neglecting  the  u^  and  w^  terms,  we  obtain  u  —  0.0031 +, 


NUMERICAL    EQUATIONS.  481 

SO  that  to  four  places  of  decimals  the  root  of  (1)  is  1.4231. 
The  process  may  evidently  be  continued  until  the  root  is 
calculated  to  any  desired  degree  of  accuracy. 

496.  We  shall  now  make  some  observations  on  the  pre- 
ceding work. 

First :  If  we  diminish  the  roots  by  a  number  less  than 
the  required  root,  as  we  do  not  pass  through  the  root,  the 
sign  of  the  last  term  remains  unchanged  throughout  the 
work.  The  last  coefficient  but  one  will  always  have  a  sign 
opposite  to  that  of  the  last  term. 

If,  in  (3),  the  signs  of  the  last  two  terms  were  alike,  the  value  of  z 
would  be  —  0.02-f-.  This  would  show  that  the  value  assumed  for  z 
was  too  great,  and  we  should  diminish  the  value  of  z  and  make  the 
last  transformation  again.  In  beginning  an  example,  one  is  very 
likely  to  assume  too  large  a  value  for  the  next  figure  of  the  root ;  in 
solving  (2),  for  instance,  the  first  solution  gave  y  =  0.5,  and  had  that 
value  been  tried,  it  would  have  proved  to  be  too  great. 

The  first  transformation  may,  however,  change  the  sign  of  the  last 
term.  Thus,  if  there  had  been  a  root  between  0  and  1  in  equation  (1), 
diminishing  the  roots  by  1  would  have  changed  the  sign  of  the  last 
term. 

Second :  In  finding  the  second  figure  of  the  root  we 
make  use  of  the  last  three  terms  of  the  first  transformed 
equation  instead  of  the  last  two  terms.  Or,  we  may  use 
the  alternative  method.  One  of  these  methods  will  gener- 
ally give  the  correct  figure.  In  any  case  we  can  find  the 
correct  figure  by  another  trial. 

Any  figure  after  the  second  is  generally  found  correctly 
from  the  last  two  terms ;  for,  in  this  case,  the  root  is  small 
and  its  square  and  cube  so  much  smaller  than  the  root 
itself  that  the  terms  in  which  they  appear  have  but  slight 
influence  upon  the  result. 

497.  It  is  not  necessary  to  write  out  the  successive  trans- 
formed equations.    When  the  coefficients  of  any  transformed 


482 


ALGEBRA. 


equation  have  been  computed,  the  next  figure  of  the  root 
may  be  found  by  dividing  the  last  coefficient  by  the  pre- 
ceding coefficient,  and  changing  the  sign  of  the  quotient. 

Thus,  in  equation  (4),  the  next  figure  of  the  root  is  obtained  by 
dividing  0.024888  by  7.9908. 

On  this  account  the  last  coefficient  but  one  of  each  trans- 
formed equation  is  called  a  trial  divisor.  ^ 

Sometimes  the  last  coefficient  but  one  in  one  of  the  transformed 
equations  is  zero.  To  find  the  next  figure  of  the  root  in  this  case 
follow  the  method  given  for  finding  the  second  figure  of  the  root. 

The  work  may  now  be  collected  and  arranged  as  follows: 


-6                      +3 
+  1                       -5 

+  5  11.423 + 

—  2 

-5 

+  1 

-2 
-4 

+  3 
-2.816 

-4 

+  1 

-6 

-1.04 

+  0.184 
-0.159112 

-3 

+  0.4 

-7.04 
-0.88 

+  0  024888 

-2.6 
+  0,4 

-2.2 
+  0.4 

-7.92 

-  0.0356 

-  7.9556 

-  0.0352 

-1.8 

+  0.02 

-1.78 
+  0.02 

-1.76 
+  0.02 

—  7  990{ 

5 

-  1.74 


NUMERICAL    EQUATIONS.  483 

The  broken  lines  mark  the  conclusion  of  each  transformation. 
The  numbers  in  heavy  type  are  the  coefficients  of  the  successive 
transformed  equations,  the  first  coefficient  of  each  equation  being  the 
same  as  the  first  coefficient  of  the  given  equation.  In  this  example 
the  first  coefficient  is  1. 

When  we  have  obtained  the  root  to  three  places  of  decimals  we 
can  generally  obtain  two  or  three  more  figures  of  the  root  by  simple 
division. 

498.  In  practice  it  is  convenient  to  avoid  the  use  of  the 
decimal  points.  We  can  do  this  as  follows :  multiply  the 
roots  of  the  first  transformed  equation  by  10,  the  roots  of 
the  second  transformed  equation  by  100,  and  so  on.  In 
the  last  example  the  first  transformed  equation  will  now  be 

%f  -  30y'  -  600y  +  3000  =  0, 

and  this  equation  will  have  a  root  between  4  and  5.     The 
second  transformed  equation  will  now  be 

2'  -  180  2^  -  79,2002  +  184,000  ^  0, 

and  this  equation  will  have  a  root  between  2  and  3.     And 
so  on. 

Comparing  these  equations  with  the  equations  in  §  495, 
we  see  that  we  can  avoid  the  use  of  the  decimal  point  by 
adopting  the  following  rule : 

When  the  coefficients  of  a  transformed  equation  have 
been  obtained,  add  one  cipher  to  the  second  coefficient,  two 
ciphers  to  the  third  coefficient,  and  so  on.  The  coefficients 
and  the  next  figure  of  the  root  are  now  integers.  The 
work  proceeds  as  in  §  497. 

If  the  root  of  the  given  equation  lay  between  0  and  1,  we  should 
^  tgin  by  multiplying  the  roots  of  the  given  equation  by  10. 


484 


ALGEBRA. 


The  complete  work  of  the  last  example,  for  six  figures  of 
the  root,  will  now  be  as  follows  : 


-180 

+     2 

-178 
+     2 

-176 
+     2 


-1740 

+ 3 

-  1737 
+ 3 

-  1734 
+        3 


- 17310 

+ 1_ 

-  17309 
+ 1 

-  17308 
+  1 


+  3 
-5 

-2 
-4 


-600 

-104 

-704 


-  79200 

-  356 

-  79556 

-  352 


7990800 

5211 

-  7996011 
5202 


800121300 

17309 

800138609 
17308 


-  800155917 


+  5    I   1.42311  + 

-2 


+  3000 

-2816 


+    184000 

-    159112 


+  24888000 

-  23988033 


+  899967000 

-  800138609 


99828391 


17307 


NUMERICAL    EQUATIONS.  485 

We  can  find  five  more  figures  of  the  root  by  simple 
division.  If  we  divide  99,828,391  by  800,155,917,  we 
obtain  0.124761,  so  that  the  required  root  to  ten  places  of 
decimals  is  1.4231124761. 

The  reason  is  seen  by  examining  the  last  transformed  equation. 
Write  this 

8.00155917^/;  =  0.000099828391  -  1.1301  w^  +  w^ 

As  w  IS  about  0.00001,  w"^  is  about  0.0000000001,  and  w^  is  still 
smaller.  Hence  the  error  in  neglecting  the  w"^  and  w^  terms  is  in  811; 
about  0.00000000017  and  in  w  about  0.00000000002.  The  result 
obtained  by  division  will  therefore  be  true  to  ten  places  of  decimals. 

499.  Negative  Koots.  To  avoid  the  inconvenience  of 
working  with  negative  numbers,  when  we  wish  to  calculate 
a  negative  root,  we  change  the  signs  of  the  roots  (§  482), 
and  calculate  the  corresponding  positive  roots  of  the  trans- 
formed equation. 

Thus  one  root  of  the  equation 

lies  between  0  and  —  1  (^  494).     By  Horner's  Method  we  find  the 
corresponding  root  of 

a;3  +  6«2  +  3«-5  =  0 

to  be  0.6696 +.     Hence,  the  required  root  of  the  given  equation  is 
-  0.6696+. 

Exercise  142. 

Compute  for  each  of  the  following  equations  the  root  of 
which  the  first  figure  is  the  number  in  parenthesis  opposite 
the  equation.  Carry  out  the  work  to  three  places  of  deci- 
mals : 

1.  x'  +  3x-^  =  0  (4-1). 

2.  a;3-6a:-12  =  0  (+3). 

3.  rr^  +  a;^  +  a;-100  =  0.  (4-4). 

4.  .T^  +  lO^c'  +  e:?;- 120-0  (+2). 

5.  x^-{-9x^-\-2^x  +  l7  =  0  (-4). 

6.  a;*-12r»-{-12a;-8  =  0  (-1). 

7.  x'-8x'  +  Ux'-^4:x-8  =  0        (-0). 


486 


ALGEBRA. 


500.  Contraction  of  Horner's  Method.  In  §  498  the  student 
will  see  that  if  we  seek  only  the  first  six  figures  of  the  root, 
the  last  six  figures  of  the  fourth  coefficient  of  the  last  trans- 
formed equation  may  be  rejected  without  affecting  the 
result.  Those  figures  of  the  second  and  third  coefficients 
which  enter  into  the  fourth  coefficient  only  in  the  rejected 
figures  may  also  be  rejected.  Moreover,  we  may  reject  all 
the  figures  which  stand  in  vertical  lines  over  the  figures 
already  rejected. 

The  work  may  now  be  conducted  as  follows : 


1        -6                         +3 
+  1                         -5 

+  5  1  1.42311+ 

-2 

-5 

+  1 

-2 
-4 

-f  yooo 

-2816 

-4 
+  1 

-600 

-104 

+    184000 
-    159112 

—  30 

+   4 

-704 
-   88 

+  24888 
-  23991 

-26 
+   4 

-  7920( 

-  35( 

) 
3 

+  897 
-800 

-22 
+   4 

-  7955( 

-  351 

3 

I 

+   97 
-   80 

—  180 

+      2 
-178 
+     2 
-176 

-  7990? 

-7991 

-  6 

-7997 
6 

J 

+          2; 

-8003 

-174 

-2 

-800 
-80 

NUMERICAL    EQUATIONS.  487 

The  double  lines  in  the  first  column  indicate  that  beyond 
this  stage  of  the  work  the  first  column  disappears  alto- 
gether. 

In  the  present  example  we  find  three  figures  of  the  root 
before  we  begin  to  contract.  We  then  contract  the  work 
as  follows : 

Instead  of  adding  ciphers  to  the  coefiicients  of  the  trans- 
formed equation,  we  leave  the  last  term  as  it  is  ;  from  the 
last  coefiicient  but  one  we  strike  off  the  last  figure ;  from 
the  last  coefiicient  but  two  we  strike  off  the  last  two  figures ; 
and  so  on.  In  each  case  we  take  for  the  remainder  the 
nearest  integer.  Thus,  in  the  first  column  of  the  preceding 
example  we  strike  ofi"  from  174  the  last  two  figures,  and 
take  for  the  remainder  2  instead  of  1. 

The  contracted  process  soon  reduces  to  simple  division. 
Thus,  in  the  last  example,  the  last  two  figures  of  the  root 
were  found  by  simply  dividing  897  by  800. 

To  insure  accuracy  in  the  last  figure,  the  last  divisor 
must  consist  of  at  least  two  figures.  Consider  the  trial 
divisor  at  any  stage  of  the  work.  If  we  begin  to  contract, 
we  strike  off  one  figure  from  the  trial  divisor  before  finding 
the  next  figure  of  the  root.  Since  the  last  divisor  is  to 
consist  of  two  figures,  the  contracted  process  will  give  us 
two  less  figures  than  there  are  figures  in  the  trial  divisor. 

Thus,  in  ^  498,  if  we  begin  to  contract  at  the  third  trial  divisor, 
—  79,908,  we  can  obtain  three  more  figures  of  the  root ;  if  we  begin 
to  contract  at  the  fourth  trial  divisor,  —  800,213,  we  can  obtain  four 
more  figures  of  the  root ;  and  so  on. 

The  student  should  carefully  compare  the  contracted 
process  on  page  486  with  the  uncontracted  on  page  484. 

501.  When  the  root  sought  is  a  large  number,  we  cannot 
find  the  successive  figures  of  its  integral  portion  by  dividing 
the   absolute   term  by  the   preceding  coefficient,   because 


488  ALGEBRA. 

the  neglect  of  the  higher  powers,  which  are  in  this  case 
large  numbers,  leads  to  serious  error. 

Let  it  be  required  to  find  one  root  of 

x'~3x'-{-llx-  4,842,624,131  -  0.  (1) 

By  trial,  we  find  that  a  root  lies  between  200  and  300.     Dimin- 
ishing the  roots  of  (1)  by  200,  we  have 

y*  +  8002/3  +  239,9972/2  +  31,998,8112/  -  3,242,741,931  =  0.      (2) 
Ify^eO,    /(2/)  =  - 273,064,071. 
If  2/ =  70,    /(2/)  =  + 471,570,139. 
The  signs  of /(2/)  show  that  a  root  lies  between  60  and  70.    Dimin- 
ishing the  roots  of  (2)  by  60,  we  obtain 

2*  +  1040 ^  +  405,597 z"  +  70,302,451  z  -  273,064,071  =  0.         (3) 
The  root  of  this  equation  is  found  by  trial  to  lie  between  3  and  4. 
Diminishing  the  roots  by  3,  we  may  find  the  remaining  figures  of  the 
root  by  the  usual  process. 

502.  Any  root  of  a  number  can  be  extracted  by  Horner's 
Method. 

Find  the -fourth  root  of  473. 

Here  a;*  =  473, 

or  a;*-f0a;3  + 0x2  + Ox -473  =  0. 

Calculating  the  root,    x  =  4.66353 -f. 

If  the  number  be  a  perfect  power,  the  root  will  be  obtained  ex- 
actly. 

503.  From  the  preceding  sections  we  obtain  the  following 
general  directions  for  solving  a  numerical  equation  : 

I.  Find  and  remove  commensurable  roots  by  §§  491-493, 
if  there  are  any  such  roots  in  the  equation. 

II.  Determine  the  situation,  and  then  the  first  figure, 
of  each  of  the  incommensurable  roots  as  in  §  494. 

III.  Calculate  the  incommensurable  roots  by  Horner's 
Method. 


NUMERICAL   EQUATIONS.  489 

Exercise  143. 

Calculate  to  six  places  of  decimals  the  positive  roots  of 
the  following  equations : 

1.  x'-Sx-l  =  0. 

2.  x'  +  2x''-4:X~^S  =  0. 

3.  3x'-f  3a;'  +  8^-32-:0. 

4.  2a;^-26a;^  + 131^ -202-0. 

5.  x'-  12a;  +  7  =  0. 

6.  x'-bx'-\-2x'-lSx  +  bb^0. 

Calculate  to  six  places  of  decimals  the  real  roots  of  the 
following  equations,  when  incommensurable : 

7.  a;3  =  35,499.  10.   :r^  =  147,008,443. 

8.  rc^=:  242,970,624.  11.    a;^  2^ +  20  =  0. 

9.  a;*  =  707,281.  12.   a;' -  lOa.-^  +  8a;+ 120  =  0. 

Sturm's  Theorem. 

604.  The  problem  of  determining  the  number  and  situa- 
tion of  the  real  roots  of  an  equation  is  completely  solved 
by  Sturm's  Theorem.  In  theory  Sturm's  method  is  per- 
fect ;  in  practice  its  application  is  long  and  tedious.  For 
this  reason,  the  situation  of  the  roots  is  in  general  more 
easily  determined  by  the  methods  already  given. 

Before  passing  on  to  Sturm's  Theorem  itself  we  shall 
prove  two  preliminary  theorems. 

505.  Sign  of  f  (x).  The  function  f'(x)  is  by  definition 
(§  476)  the  limit  of  the  ratio  of  the  increment  off(x)  to  the 
corresponding  increment  of  x,  as  the  latter  approaches  0  as 
a  limit. 


490 


ALGEBRA. 


If  we  suppose  the  increment  of  x  to  be  always  positive, 
then  the  corresponding  increment  of  f(x)  may  be  positive 
or  negative.  If  the  increment  of  f(x)  is  positive,  however 
small  the  increment  of  x  may  be,  then  in  the  limit  f'(x) 
will  be  positive.  But  if  for  very  small  increments  of  x, 
the  increment  oif{x)  is  always  negative,  then /'(a;)  will  be 
negative. 

In  other  words,  f  (x)  is  positive  or  negative  for  any  value 
of  X  according  as  the  function  f  (x)  is  increasing  or  decreas- 
ing as  X  increases  from  this  particular  value  ofx. 

Referring  to  the  graph  of  f(x),  f'(x)  will  be  positive  as 
long  as  the  curve  is  rising  toward  the  right,  and  negative 

when  the  curve 
is  sinking  toward 
the  right.  At  the 
highest  and  low- 
est points,  £  and 
A,  the  derivative 
changes  its  sign, 
that  is,  it  passes 
through  the  value 
0  at  these  points.  The  values  of  x  corresponding  to  A  and 
B  are  roots  of  the  equation  f(x)  =  0. 

506.  Signs  of  f  (x)  and  f  (x).  Let  a  he  any  real  root  of  an 
equation  f  (x)  =  0,  which  has  no  equal  roots. 

Let  X  change  continuously  from  a  —  h,  a  value  a  little  less 
than  a,  to  a-j-h,  a  value  a  little  greater  than  a.  Then  f  (x) 
and  f  (x)  will  have  unlike  signs  immediately  before  x  passes 
through  the  root,  and  like  signs  immediately  after  x  passes 
through  the  root. 

For,  in  the  graph,  if  f(x)  is  positive  just  before  x  passes 
through  a  root  as  at  F  and  H,  the  curve  is  sinking  to  the 
right,  and  therefore  f(x)  is  negative,  both  before  and  after 


NUMERICAL    EQUATIONS.  491 

the  root  is  passed.  But /(a;)  is  positive  before  we  reach  the 
root  and  negative  after  we  pass  it. 

Again  if  f(x)  is  negative  just  before  we  reach  a  root,  as 
at  Q,  the  curve  is  rising  to  the  right,  and  therefore /' (a;)  is 
positive  on  both  sides  of  the  root,  while  f(x)  changes  its 
sign  from  —  to  +  as  we  pass  through  Q. 

Hence,  in  both  cases /(a;)  and /'(a;)  have  unlike  signs  just 
before  we  reach  a  root,  and  like  signs  as  soon  as  we  have 
passed  a  root. 

507.  Sturm's  Functions.  The  process  of  finding  the  H.  C.  F. 
off(x)  and /'(a:)  has  been  employed  (§  480)  in  obtaining 
the  multiple  roots  of  the  equation  f(x)  =  0.  We  use  the 
same  process  in  Sturm's  Method. 

Let  f(x)  —  0  be  an  equation  which  has  no  multiple 
roots  ;  let  the  operation  of  finding  the  H.  C.  F.  of  f(x)  and 
f'(x)  be  carried  on  until  the  remainder  does  not  involve  x, 
the  sign  of  each  remainder  obtained  being  changed  before  it 
is  used  as  a  divisor. 

If  there  is  a  H.C.  F.,  the  equation  has  multiple  roots.  Remove 
them  and  proceed  with  the  reduced  equation. 

Kepresent  hj  fiix),  f3{x), fn{x)  the  several  remain- 
ders with  their  signs  changed.  These  expressions  with 
f'(x)  are  called  Sturm's  Functions. 

Now,  if  D  represents  the  dividend,  d  the  divisor,  q  the 
quotient,  and  H  the  remainder, 

I)  =  qd-}-Ii. 
Consequently,    /  (x)  =  qif  (x)  —  f^  (:?;), 
f\x)  =  q,f,{x)-f,(xl 
fi{x)  =  q3fs(x)-fi(x), 


fn-2  (X)  =  2'n-l/«-l  W  -/,  (^)  ; 


492  ALGEBRA. 

where  ^i,  q-i, g„_i  represent  the  several  quotients,  or  the 

quotients  multiplied  by  positive  integers. 

From  these  identities  we  have  the  following  , 

I.  Two  consecutive  functions  cannot  vanish  for  the  same 
value  of  X. 

For  example,  suppose /^  (:r)  and /a  (a;)  to  vanish  for  a  par- 
ticular value  of  X.  Give  to  x  this  value  in  all  the  identi- 
ties. By  the  third  identity,  fi  {x)  will  vanish ;  by  the 
fourth,  /s  {x)  will  vanish  ;  finally,  /„  (x)  will  vanish  ;  but 
this  is  contrary  to  the  hypothesis  that  f{x)  =  0  has  no 
multiple  roots. 

II.  When  we  give  to  a;  a  value  which  causes  any  one 
function  to  vanish,  the  adjacent  functions  have  opposite 
signs. 

Thus,  if/3(a;)  =  0,  from  the  third  identity /2 (:r)  =  —fi(x). 

508.  Sturm's  Theorem.  We  are  now  in  a  position  to  enun- 
ciate Sturm's  Theorem : 

If  in  the  series  of  functions 

f(x),   f'(l),   f,(x) f„(3j) 

tue  give  to  x  any  particular  value  a,  and  determine  the  num- 
ber of  variations  of  sign ;  then  give  to  x  any  greater  value  b, 
and  determine  the  number  of  variations  of  sign ;  the  number 
of  variations  lost  is  the  number  of  real  roots  of  the  equation 
f  (x)  =  0  betweeii  a  and  b. 

Let  x  increase  continuously  from  a  to  b. 

First :  Take  the  case  in  which  x  passes  through  a  root  of 

any  of  the  functions  f{x),  f{x), /„_i(a;),  for  example, 

fi{x).  The  adjacent  functions  in  this  case  have  opposite 
signs.    f{x)  itself  changes  sign,  but  this  has  no  effect  on 


NUMERICAL    EQUATIONS.  493 

the  number  of  variations ;  for  if  just  before  x  passes  through 

the  root  the  signs  are  +  H ,  just  after  x  passes  through 

the  root  they  will  be  H ,  and  the  number  of  variations 

is  in  each  case  one. 

Hence,  there  is  no  change  in  the  number  of  variations  of 
sign  when  x  passes  through  a  root  of  any  of  the  functions 
f{x\f,{x\ fn-xix). 

Second :  Take  the  case  in  which  x  passes  through  a  root 
oi  f{x)  =  0.  Since  f{x)  and  /'(^)  have  unlike  signs  just 
before  x  passes  through  the  root,  and  like  signs  just  after 
(§  506),  there  is  one  variation  lost  for  each  root  oif{x)  =  0. 

Hence,  the  number  of  real  roots  between  a  and  h  is  the 
number  of  variations  of  sign  lost  as  x  passes  from  a  to  h. 

To  find  the  total  number  of  real  roots,  we  take  x  first 
very  large  and  negative,  and  then  very  large  and  positive. 
The  sign  of  each  function  is  then  the  sign  of  its  first  term. 

The  student  may  not  understand  how  it  is  that/(a;)  and/'(a;)  always 
have  unlike  signs  just  before  x  passes  through  a  root. 

Let  o  and  fi  be  two  consecutive  roots  of  f{x)  =  0  ;  let  ^  be  very 
small.  Suppose  that  f{x)  changes  at  a  from  +  to  —  ;  then  f{a)  is  — 
(g  505). 

When  x  =  a  —  h,    /(x)  = +,  /(re)  is  -  ; 

a;  =  a,  m  =  0,  fix)i&-. 

As  X  changes  from  o  to  $,f{x)  passes  through  an  odd  number  of 
roots  (§  494),  and  consequently  changes  sign.  Hence,  when  x  =  $—h, 
f{x)  is  — ,  f{x)  is  +  ;  and /(a;)  and  f{x)  again  have  unlike  signs. 

509.  Example.  Determine  the  number  and  signs  of  the 
real  roots  of  the  equation 

x'  —  4a:'  +  6a:^  —  12a;  +  1  =  0. 

Here  f{x)  =  4  x^  -  12  x^  +  12  x  -  12. 


494 


ALGEBRA. 


Let  us  take  ioiJ\x),  however,  the  simpler  expression 

"We  proceed  as  if  to  find  the  H.  C.  F.,  changing  the  sign  of  each 
remainder  before  using  it  as  a  divisor. 


1_   3+   3_  3 

3-   9+    9-   9 
3+    1 


-10 

+    9 

-30 

+  27 

-30 

-10 

37- 

-   9 

111- 

27 

111 -f 

37 

-64 
+  64 


1-4  +  6-12  +  1 

l_3+3_    3 


-1  +  3 
-1  +  3 


9  +  1 
3  +  3 


-    6-2 
3  +  1 


1-1 


1-10  +  37 


The  coefficients  of  the  several  functions  are  in  heavy  type.  In 
the  ordinary  process  of  finding  the  highest  common  factor  we  can 
change  signs  at  pleasure.  In  finding  Sturm's  functions  we  cannot 
do  this  as  the  sign  is  all  important.  We  can,  however,  take  out  any 
positive  factor. 

We  now  have  f{x)  =  a;*  —  4x^  +  6a;2  —  12x  +  1, 

f{x)^a^-3x'  +  3x-3, 

f,{x)  =  3x  +  l, 

Mx)  =  +  6i. 
When  fix)    fix)    Mx)   /.{x) 


x  =  -1000 

+ 

- 

- 

+ 

2  variations. 

x=     0 

+ 

— 

+ 

+ 

2  variations 

X  =  +  1000 

+ 

+ 

+ 

+ 

0  variations 

Hence  the  equation  has  two  real  positive  roots  ;  it  must  therefore 
have  two  imaginary  roots. 

The  real  roots  will  be  found  by  §  494  to  lie  one  between  0  and  1, 
and  one  between  3  and  4. 


NUMERICAL   EQUATIONS.  495 

Exercise  144. 

Determine  by  Sturm's  Theorem  the  number  and  situation 
of  the  real  roots  of  the  following  equations  : 

1.  a;' -4a;' -11a:  4- 43  =  0. 

2.  a^-6x''  +  7x-3  =  0. 

3.  x*-4:x'-\-x''-{-e>x-\-2  =  0. 

4.  a:*-5a;'-fl0a;'^-6a:-21=0. 

5.  a;* -a;' -a;'' 4- 6  =  0. 

6.  a;* -2a;'- 3a;' +  10a;- 4  =  0. 

7.  a;^  +  2a;*+3a;'  +  3a;'-l  =  0. 

8.  a;5  +  a;'-2a;'  +  3a;-2  =  0. 

610.  The  Cube  Roots  of  Unity.  The  equation  a;'  =  1,  or 
a;'  —  1  =  0  may  be  written 

(a;-l)(a;'  +  ^+l)-=0. 

The  roots  are     1,  -  i  +  iV^  -  i  -  -J-V^^. 

If  either  of  the  imaginary  roots  is  represented  by  w,  the 
other  is  found  by  actual  multiplication  to  be  a>' ;  also 

a>'  +  0)  +  1  =  0. 

Every  number  u  has  three  cube  roots.  If  one  of  its  roots 
be  u^,  the  others  are  <i>u^  and  wV*.  For  the  cube  of  any 
one  of  these  is  evidently  u. 

511.  The  General  Oubic.  We  shall  write  the  general 
equation  of  the  third  degree  in  the  form 

aa;'  +  3  ^>a;'  +  3  ca;  +  c^  =  0.  (1) 


496  ALGEBRA. 

Before    attempting   to    solve  this  we  shall  transform  it 

into  an  equation  in  which  the  second  term  is  wanting. 

z  —  h 

Put   z  —  ax-\-h]    :.  x  = .      Substituting   this   ex- 

a 

pression  for  x,  and  reducing,  we  obtain 

or,  putting   11=  ac  —  &^,  G  =  a^d  —  3  ahc  +  2  5^, 

z'-\-2.IIz-\-G  =  0,  (2) 

in  (2)  put  z  =  u^  -\-  v^. 

Then 

{u^  +  v^f  +  3  II{J  +  v^)  +  6^  =  0, 
which  reduces  to 

.      w  +  'y  +  3(wM  +  ir)(w^  +  'y*)+6^=-0.  (3) 

Since  we  have  assumed  but  one  relation  between  u  and 
V,  we  can  assume  one  more  relation.     Let  us  assume 

uK^  =  ~H.  (4) 

Equation  (3)  now  reduces  io  U'\-v  =  —  O.  (5) 

And  (4)  may  be  written  uv  =  —  H^.  (6) 

Eliminating  v,  we  obtain  the  quadratic 

u"  +  Gu  =  IP,  (7) 

called  the  reducing  quadratic  of  the  cubic. 
Solving  this  quadratic,  we  find 


(8) 


-G± 

^G'  +  ^H' 

2 

-IP 

-G^-\/G'-V^.H' 

u 

2 

NUMERICAL    EQUATIONS.  497 

Since  u  and  v  differ  only  in  having  opposite  signs  before 
the  radical,  it  will  make  no  difference  in   the   equation 

ax -\- h  ^^  z  =  u^  -{- v^  whether  the  radical  be  taken  with 
the  +  sign  in  u  or  in  v. 

Again,  when  any  one  of  the  three  values  u^  has  been 
selected,  the  corresponding  value  of  v^  must  be  so  taken 
that  (4)  is  satisfied.  Accordingly  we  obtain  just  three 
solutions  for  z,  and  consequently  for  x,  corresponding  to  the 
three  values  of  u^.     The  three  values  of  z  are 

i  ^  k  ^  2     1  ^ 

where  u'^  is  any  one  of  the  three  cube  roots  of  u. 
The  above  solution  is  known  as  Cardans. 

Solve,  by  Cardan's  method. 

Here  a  =  3,  6  =  4.     Putting  2  =-  3  or  +  4,  we  obtain 
2' -122 -34  =  0, 
•.  ^=-4,  G'  =  -34, 
and  the  reducing  quadratic  is  ^ 

w2_34w  =  -64. 
Solving,  w  =  2  or  32  ; 

.-.  V  =  -  —  =  32  or  2. 

u 

Hence  the  values  of  2  are 

\/2+v'32,    «v/2  +  ft,2^^32_    „'v/2  +  «v^; 
or  v'2  +  2^,  a,v^2  +  2ft,2v/4.   «2^2  +  2«v^4; 

and  the  values  of  x  are 
i(\/2 +  2v^4-4),   ^0.^2  +  20,2^4- 4),    |(o,'v/2  f  2o,v/4 -4). 


498  ,  ALGEBRA. 

512.  Discussion  of  the  Solution.  Cardan's  method  furnishes 
a  complete  algebraic  solution  of  the  equation  of  the  third 
degree,  but  is  of  little  value  in  solving  nuvierica<l  equations. 

It  may  be  shown  that  the  expression  G'^  +  4^'  is  nega- 
tive, zero,  or  positive,  according  as  the  three  roots  of  the 
given  equation  are  real  and  unequal,  two  of  them  equal, 
or  two  imaginary  and  one  real.  The  expression  G"^  +  AIT^, 
whose  value  determines  the  nature  of  the  roots,  is  called 
the  Discriminant  of  the  cubic.     Consider  the  three  cases. 

I.  All  three  roots  real  and  unequal.  In  this  case, 
G^-{-4:ir^  is  negative,  and  its  square  root  is  imaginary. 
If  we  put  JT^  =  -  ( (?'  +  4  ^),  we  have 

Since  there  is  no  general  algebraic  rule  for  extracting 
the  cube  root  of  an  imaginary  expression,  the  case  of  three 
real  and  unequal  roots  is  known  as  the  irreducible  case. 

II.  Tivo  of  the  roots  equal.     Then   (9^+4^  =  0,  and 

G^ 


ax  -]-  b 


ax  +  b^-(z^^  + 


o 


III.    Tlvo  roots  imaginary.     Then  6^^^  +  4  ^^  is  positive, 
its  square  root  is  real,  and  we  have 


Hence,  the  general  solution  gives  us  the  roots  of  a  numer- 
ical cubic  in  a  form  in  which  their  values  can  be  readily 
computed  only  in  the  second  and  third  cases. 

In  the  first  case  the  roots  may  be  calculated  by  Trigo- 
nometry.    (See  Wentworth's  College  Algebra,  §  635.) 

The  real  roots,  however,  are  more  easily  found  by 
Horner's  method. 


CHAPTER  XXXVII. 

DETERMINANTS. 

513.   Origin.     Solving  the  two  simultaneous  equations 

aiX  +  biy  =  Ci, 

a^x  +  b,7/  -■=  Cj, 


we  obtain 


Similarly,  from  the  three  simultaneous  equations 
ayx  +  %  4-  CyZ  =  c?i, 

a^x  +  h.,y  4-  C32;  ---  4, 
we  obtain 

_^  c/i^agg  —  dyhjCj  4-  cZ^^ggi  —  C?2^ig3  4-  dJ)^C2  —  C?352^l 

«1^2<^3  —  (^\i>3p2  4-  «2^3<^1  —  «2^1^3  4-  «3^1<?2  —  a3^2^1  ' 

with  similar  expressions  for  y  and  z. 

The  numerators  and  denominators  of  these  fractions  are 
examples  of  expressions  which  often  occur  in  algebraic 
work,  and  for  which  it  is  therefore  convenient  to  have  a 
special  name  ;  such  expressions  are  called  determinants. 

514,  Definitions.  Determinants  are  usually  written  in  a 
compact  form,  called  the  square  form. 

Thus,  a^\  -  a^\  is  written        ^       ^ 


500  ALGEBRA. 

and  «i^2^3  ~  ^1^3'^2  +  ^'l^-A^\  ~  <^2^1^3  +  <^3'^^lC2  —  «3^2^1 

is  written  61       h^      I 

This  square  form  is  sometimes  ^ritteH-i^a  still  more  abbreviated 
form.  Thus,  the  last  two  determinants  are  written  \  a^  W  and 
!«!  62  <^3l-  1'^is  last  notation  should,  however,  always  suggest  the 
square  form ;  in  any  problem  it  will  generally  be  advisable  to  write 
out  this  abbreviated  form  in  the  complete  square  form. 

The  individual  symbols  ai,  a^,  bi,  b^,  ,  are  called  ele- 
ments. 

A  horizontal  line  of  elements  is  called  a  row;  a  vertical 
line  a  column. 

The  two  lines  aj,  Jj,  <?3  and  a-i,  b.^,  c^  are  called  diagonals  ; 
the  first  the  principal  diagonal,  the  second  the  secondary 
diagonal. 

The  order  of  a  determinant  is  the  number  of  elements  in 
a  row  or  column. 

Thus,  the  last  two  determinants  are  of  the  second  and  third  orders, 
respectively. 

The  expression  of  which  the  square  form  is  an  abbrevia- 
tion is  called  the  expanded  form,  or  simply  the  expansion,  of 
the  determinant. 

The  several  terms  of  the  expansion  are  called  terms  of 
the  determinant. 


Thus  the  expansion  of 


is   a^h^  —  aj)i. 


Remark.     By  some  writers  constituent  is  used  where  we  use  ele-  • 
ment,  and  element  where  we  use  term. 

515.  General  Definition.  In  general,  a  determinant  of  the 
nih.  order  is  an  expression  involving  ?z^  elements  arranged 
in  n  rows  of  n  elements  each ;  the  expansion,  that  is,  the 
expression  for  which  the  square  form  is  an  abbreviation, 
being  found  as  follows : 


DETERMINANTS.  501 

Form  all  the  possible  products  of  n  elements  each  that 
can  be  formed  by  taking  one,  and  only  one,  element  from 
each  row,  and  one,  and  only  one,  element  from  each  col- 
umn ;  prefix  to  each  of  the  products  thus  formed  either  + 
or  —  (which  sign  is  to  be  determined  by  a  rule  to  be  given 
in  the  following  sections),  and  take  the  sum  of  all  these 
products. 

Note.  Nearly  all  the  properties  of  determinants  can  be  obtained 
directly  from  this  definition  and  the  rule  of  signs  (§  518  or  §  519). 
This  will  be  the  method  followed  in  the  present  chapter.  It  is  there- 
fore of  the  utmost  importance  that  the  student  should  thoroughly 
understand  the  present  and  the  four  following  sections. 

516.  Inversions  of  Order.  In  any  particular  determinant 
the  letters  and  subscripts  in  the  principal  diagonal  are  said 
to  be  in  the  natural  order.  If  the  letters,  of  subscripts,  are 
taken  in  any  other  order,  there  wnll  be  one  or  more  invei^- 
sions  of  order. 

Thus,  if  1,  2,  3,  4,  5  be  the  natural  order,  in  the  order  2,  3,  5,  1,  4, 
there  will  be  four  inversions :  2  before  1,  3  before  1,  5  before  1,  5  be- 
fore 4. 

Similarly,  if  a,  b,  c,  d  be  the  natural  order,  in  the  order  h,  d,  a,  c, 
there  will  be  three  inversions :  b  before  a,  d  before  a,  d  before  c. 

517.  In  any  series  of  integers  (or  letters)  let  two  adjacent 
integers  (or  letters)  be  interchanged ;  then,  the  number  of 
inversions  is  either  increased  or  diminished  by  one. 

For  example,  in  the  series  6  2  [5  1]  4  3  7,  interchange  5  and  1. 

We  now  have  6  2  [1  5]  4  3  7. 

The  inversions  of  5  and  1  with  the  integers  before  the  group  are 
the  same  in  both  series. 

The  inversions  of  5  and  1  with  the  integers  after  the  group  are 
the  same  in  both  series. 

In  the  first  series  5  1  is  an  inversion  ;  in  the  second  series  1  5  is  not. 

Hence,  the  interchanging  of  5  and  1  diminishes  the  number  of 
inversions  by  one. 

Similarly,  for  any  case. 


502  ALGEBRA. 

518.  Signs  of  the  Terms.  The  principal  diagonal  term 
always  has  a  +  sign. 

To  find  the  sign  of  any  other  term :  Add  together  the 
number  of  inversions  among  the  letters,  and  the  number  of 
inversions  among  the  subscripts.  If  the  total  number  is 
even,  the  sign  of  the  term  is  -}-  ;  if  odd,  —. 

Thus,  in  the  determinant  ja^  b^  c^  d^\  consider  the  term  c^a^d^b^. 
There  are  in  c  a  db  three  inversions ;  in  2  3  4  1  three  inversions  ;  the 
total  IS  six,  an  even  number,  and  the  sign  of  the  term  is  +. 

519.  In  practice  the  sign  of  a  term  is  easily  found  by 
one  of  the  following  special  rules: 

Rule  I.  Write  the  elements  of  the  term  in  the  natural 
order  of  letters ;  if  the  number  of  inversions  among  the  sub- 
scripts is  even,  the  sign  of  the  term  is  -\- ;  if  odd,  — . 

Rule  II.  Write  the  elements  in  the  natural  order  of  sub- 
scripts ;  if  the  numher  of  inversions  among  the  letters  is  even, 
the  sign  of  the  term  is -^ ;  if  odd,  —. 

Thus,  in  the  determinant  \a^  b^  c^  d^\  consider  the  term,  c^a^dj)^. 
Writing  the  elements  in  the  order  of  letters,  we  have  a^b^c,^d^. 
There  are  two  inversions,  viz.  :  3  before  1,  and  3  before  2;  and  the 
sign  of  the  term  is  +.  Or,  write  the  elements  in  the  order  of  sub- 
scripts, b^c^a^d^.  There  are  two  inversions,  viz;:  b  before  a,  and  c 
before  a  ;  and  the  sign  of  the  term  is  +. 

That  these  special  rules  give  the  same  sign  as  the  general  rule  of 
§  518  may  be  seen  as  follows : 

Consider  the  term  c^a^dj)^.  Its  sign  is  determined  by  the  total 
number  of  inversions  in  the  two  series  03  4  i-  Bring  a^  to  the  first 
position  ;  this  interchanges  in  the  two  series  c  and  a,  2  and  3.  In 
each  series  the  number  of  inversions  is  increased  or  diminished  by 
one  (^  517),  and  the  total  is  therefore  increased  or  diminished  by  an 
even  number. 

Interchange  b^  and  d^,  then  interchange  b^  and  c^ ;  this  brings  b^  to 
the  second  place,  and  the  letters  into  the  natural  order.  As  before, 
the  total  number  of  inversions  is  changed  by  an  even  number. 


DETERMINANTS. 


503 


and  the  number  of  inversions 
differs  by  an  even  number  from  that  found  by  the  general  rule  of 
g  518.  Hence,  the  sign  given  by  Rule  I.  agrees  with  the  sign  given 
by  the  general  rule. 

520.  If  all  the  elements  in  any  row  (or  column)  are  zero, 
the  determinant  is  zero.  For  every  term  contains  one  of 
the  zeros  from  this  row  (or  column)  (§  515),  and  therefore 
every  term  of  the  determinant  is  zero. 

A  determinant  is  unchanged  if  the  rows  are  changed  to 
columns  and  the  columns  to  rows.  For  the  rules  (§§  515, 
518)  are  unchanged  if  "  row  "  is  changed  to  "  column  "  and 
"  column  "  to  "row." 


Thus, 


h     K 


«3 

«! 

^ 

h' 

■= 

a., 

h 

^3 

«3 

h 

521.    A  determinant  of  the  third  order  may  be  conven- 
iently expanded  as  follows : 


Three  elements  connected  by  a  full  line  form  a  positive 
term ;  three  elements  connected  by  a  dotted  line  form  a 
negative  term.    The  expansion  obtained  from  the  diagram  is 

aib^c^  +  «2^3^i  +  «3^i^2  —  «i^3^2  —  a-i^iCs  —  a^b-iCi, 
which  agrees  with  §  518. 

There  is  no  simple  rule  for  expanding  determinants  of 
orders  higher  than  the  third. 


504 


ALGEBRA. 


Exercise  145. 
Prove  the  following  relations  by  expanding 

1. 


2. 


«! 

a. 

«! 

h 

aj 

(h 

i. 

i. 

i. 

h 

(h 

h 

h 

J, 

(h 

ai 

«! 

az 

«3 

^1 

h 

^a 

=^ 

Ci 

C-i 

C-i 

as      tta      <^i 

^3         ^2         ^1 
^3         ^2         ^1 


5i  Ci  «! 
^3  ^3  «3 
^a        ^2        «2 


Find  the  values  of: 


3. 


1     2    3 

3     2     4 

2    4    4 

4. 

7     6     1 

5. 

3    4     5 

5     3     8 

6.    Count  the  inversions  in  the  series 
5413  2.  751436  2. 

4152  3.  654213  7. 


4       5      2 

-1       2-3 

6-4       5 

d  a  c  e  b. 
c  e  b  d  a. 


7.    In  the  determinant  |  a^  b^  c^  d^,  e^  \  find  the   signs  of 
the  following  terms: 

ajy^c^d^e^.  a^^c^d^e^.  .         exC^<i.^-^d^. 


aJy-oCz 


b^c^aie-id-i. 


Cxa^-ie^d^. 


8.  Write,  with  their  proper  signs,  all  the  terms  of  the 
determinant  |  a^  b.^  c^d^^ 

9.  Write  with  their  proper  signs,  all  the  terms  of  the 
determinant  |  a^  b^  c^  di,  e^  \  which  contain  both  aj  and  bi\  all 
the  terms  which  contain  both  b^  and  e^. 

Expand  the  determinants : 


10. 


ab  0  0 

0  0  0a 

a  b  c  0 

5  a  0  0 
0  a  a  b 

11. 

0  0  5  0 
a  a  b  b 

12. 

c  a  b  0 
b  c  a  0 

0  b  b  a 

b   h  a  a 

a  b  c  1 

DETERMINANTS. 


505 


522.  Number  of  Terms.  Consider  a  determinant  of  the 
nth  order. 

In  forming  a  term  we  can  take  from  the  first  row  any- 
one of  n  elements ;  from  the  second  row  any  one  oi  n  —  1 
elements  ;  and  so  on.  From  the  last  row  we  can  take  only 
the  one  remaining  element. 

Hence,  the  full  number  of  terms  is  n(n  —  1) 1,  or  |n. 


523.  Interchange  of  Columns  (or  Eows).  J^  two  adjacent 
columns  of  a  determinayit  A  are  interchanged,  the  determi- 
nant thus  obtained  w  —  A. 

For  example,  consider  the  determinants 


ftl 

a. 

«3 

ai 

h. 

h 

h 

h 

,        A'  = 

G\ 

C2 

Cz 

Oi 

d. 

d. 

d. 

d. 

ai 

Oz 

a^ 

at 

h 

bs 

h 

*. 

Cx 

Cz 

Cj 

<:* 

d. 

d. 

d. 

d. 

A  = 


The  individual  elements  in  any  row  or  column  of  A'  are 
the  same  as  those  of  some  row  or  column  of  A,  the  only 
difference  being  in  the  arrangement  of  elements.  Since 
every  term  of  each  determinant  contains  one,  and  only 
one,  element  from  each  row  and  column,  every  term  of  A' 
must,  disregarding  the  sign,  be  a  term  of  A. 

Now  the  sign  of  any  particular  term  of  A'  is  found  from 
a  series  (§  519,  Rule  I.)  in  which  3  2  is  the  natural  order. 
The  sign  of  the  term  of  A  which  contains  the  same  elements 
is  found  from  a  series  in  which  3  2  is  regarded  as  an  inver- 
sion. Consequently  every  term  which  in  A'  has  a  +  sign 
has  in  A  a  —  sign,  and  vice  versa  (§  517). 


Therefore 


A'=-A. 


Similarly  if  any  two  adjacent  columns  or  rows  of  any 
determinant  are  interchanged. 


506  ALGEBRA. 

524.  In  any  determinant  ^^  if  a  particular  column  is 
carried  over  m  columns,  the  determinant  obtained  is  (—  1)'"A. 

For,  successively  interchange  the  column  in  question 
with  the  adjacent  column  until  it  occupies  the  desired  posi- 
tion. There  will  be  m  interchanges  made,  and  since  there 
will  be  m  changes  of  sign  (§  523),  the  new  determinant 
will  be  (-  1)'"A. 

Similarly  for  a  particular  row. 

525.  In  any  determinant  A  if  any  two  columns  are  inter- 
changed, the  determinant  thus  obtained  zs  —  A. 

Let  there  be  m  columns  between  the  columns  in  question. 

Bring  the  second  column  before  the  first.  The  second 
column  will  be  carried  over  w  -f  1  columns,  and  the  deter- 
minant obtained  is  (-  1)'"+^A  (§  524). 

Bring  the  first  column  to  the  original  position  of  the 
second.  The  first  column  will  be  carried  over  m  columns, 
and  the  determinant  obtained  is  (—  !)"*(—  1)'"+^A,  or 
(-  1)2'"+^A. 

Since  2m-\-\  is  always  an  odd  number,  this  is  —  A. 

Similarly  for  two  rows. 

Thus, 


«1 

«2 

«3 

«3 

«2 

«1 

«3 

«2 

«1 

h 

\ 

^3 

=  - 

^3 

h 

h 

=       ^3 

C2 

Cl 

Cl 

c, 

Cs 

Cs 

^2 

Cl 

^^3 

\ 

h 

526.  Useful  Properties.  If  two  columns  of  a  determinant 
are  identical,  the  determinant  vanishes. 

For,  let  A  represent  the  determinant. 

Interchanging  the  two  identical  columns  ought  to  change 
A  into  —  A.  But  since  the  two  columns  are  identical,  the 
determinant  is  unchanged. 

.-.  A  =  - A,    2A=0,     A=0. 

Similarly,  if  two  rows  are  identical. 


DETERMINANTS. 


507 


527.  If  all  the  elements  in  any  column  he  inultiplied  by 
any  number  m,  the  dete^'minant  ivill  be  multiplied  by  m. 

For  every  term  contains  one,  and  only  one,  element  from 
the  column  in  question.  Hence  every  term,  and  conse- 
quently the  whole  determinant,  is  multiplied  by  m.- 

Similarly  for  a  row. 


Thus, 


Again, 


mai 

«2         «3 

«i 

mbi 

K      h 

=  m 

&i 

mc^ 

c,       c. 

Cl 

be 

a      a'  1 

abc 

abc 

ca 

b      b^ 

bea 

ah 

C         6-2 

cab 

6J  = 


ina^ 

-mbi 

mq 

«2 

h 

C-i 

«3 

h 

H 

1 

a2 

a^ 

= 

1 

62 

¥ 

1 

c^ 

(? 

ai  -{-  a      a^ 

«3 

bi  +  (i     Ih 

^3 

ci-^y     c., 

^3 

a 

«2 

«3 

iS 

h 

h^ 

y 

C-i 

C3 

528.  If  each  of  the  elements  in  a  column  is  the  sum  of 
two  numbers,  the  determinant  may  be  expressed  as  the 
sum  of  two  determinants. 

ai     a.2     a^ 
Thus.     br-h3     Ih     b,  =  hi     b,     b,  + 

Ci       C2       C3 

For,  consider  any  term,  as  (%  +  a)  h^Cs.  This  may  be 
written  ai^2<^3  +  ab^Ci.  Hence,  every  term  of  the  first  de- 
terminant is  the  sum  of  a  term  of  the  second  determinant 
and  a  term  of  the  third  determinant.  Consequently  the 
first  determinant  is  the  sum  of  the  other  two  determinants. 

Similarly  for  any  other  case. 

529.  If  the  elements  in  any  column  (or  row)  are  multi- 
plied by  any  number  m,  and  added  to,  or  subtracted  from, 
the  corresponding  elements  in  any  other  column  (or  row), 
the  determinant  is  unchanged. 


Thus, 


«!  =b  ma-i 

a., 

«3 

Cli 

a^    a., 

ma^ 

a-i 

«3 

hx  ±1  mbi 

h 

^3 

=. 

b. 

h,    b. 

±2 

mb-i 

h 

^3 

Cj  ifc  mc-i 

c% 

Cz 

Ci 

Ci     c. 

mc2 

C'X 

Cz 

508 


ALGEBRA. 


The  last  determinant  may  be  written 


m 


bi    b^ 


,  and  therefore  vanishes  (§  526). 


Hence,  we  have  only  the  first  determinant  on  the  right- 
hand  side. 

Similarly  for  any  other  case. 

This  process  may  be  applied  simultaneously  to  two  or 
more  columns  (or  rows)  ;  but  in  this  case  care  must  be 
taken  not  to  make  two  columns  (or  rows)  identical  (§  526). 

The  last  property  is  of  great  use  in  reducing  determi- 
nants to  simpler  forms. 


530. 

Examples. 

(1) 

b  -\-  c     a     1 
c-^a    b     1 
a-{-b     c     \ 

b-^c  +  a 
c  +  a+  b 
a^b-{-c 

{a  +  b  +  c) 


by  adding  the  second  column  to  the  first. 


a     1 

b     1 

c     1 

1    a 

1 

1     b 

1 

1     c 

1 

(2) 


14     15     11 

3 

21     22     16 

=z 

5 

23    29     17 

6 

= 

2 

11 

16 


12     17 

3     2 

5  3 

6  6- 


=  2 


3     2 

11 

5     3 

16 

6     6 

17 

2  (19) -38. 


Begin  by  subtracting  the  third  column  from  the  first  and  second 
columns.  Then  take  out  the  factor  2,  subtract  3  times  the  first  col- 
umn from  the  third,  and  multiply  out  the  result  by  ^  521. 


DETERMINANTS. 


509 


Exercise  146. 


Show  that 


1. 


0 

a 

b 

a 

0 

c 

b 

c 

0 

a 

b 

a 

b 

a 

a 

b 

a 

h 

3. 


2abc.     2. 


b  -\-  c        a  a 

b        c  -\-  a        b 
c  c        a-\-b 


=  -{a-bf{a-\-b). 


4  abc. 


1  a  d^ 
1  b  b' 
1     c     c' 


(a  —  b)(b  —  c)(c  —  a). 


Find  the  values  of 


3    5     7 

2     1     3 

.       6. 

4     3     7 

2 

13 

20 

3 

9 

18 

7. 

5 

10 

23 

19     13 

16 

25     16 

28 

28     10 

19 

Show  that 

8. 

a     a^     be 
h     b'     ac 
c     c^     ah 

=  —  ( 

9. 

a  +  2b     a  +  46 
a+35     a  +  55 
a  +  46     a  +  6b 

0. 

(a  +  by 

a'          ( 
b' 

.^  +  0) 

b' 

(a  —  b)(b  —  c)(e  —  a)(ab  +  be  +  ca). 


a  +  6b 

a  +  lb 

= 

a  +  Sb 

c' 

a' 

(c  +  a 

y\ 

0. 


2abc(a-{-b-{-cy. 


531.  Minors.  If  one  row  and  one  column  of  a  determi- 
nant be  erased,  a  new  determinant  of  order  one  lower  than 
the  given  determinant  is  obtained.  This  determinant  is 
called  a  first  minor  of  the  given  determinant. 


610 


ALGEBRA. 


Similarly,  by  erasing  two  rows  and  two  columns  we  ob- 
tain a  second  minor;  and  so  on. 

Thus,  in  the  determinant  {a^  h^  Cg],  erasing  the 
second  row  and  third  column,  we  obtain  the  first 


«i 


This  minor  is  said  to  correspond  to 


«3 


the  element  Jg,  and  is  generally  represented  by  Aj^ ; 
case,  A»3  =  1^1     ^2 1 


that. 


thi 


In  general,  to  every  element  corresponds  a  first  minor 
obtained  by  erasing  the  row  and  column  in  which  the  given 
element  stands. 


532.   Theorem.     IJf  all  the  elements  of  the  first  row  aftei- 
the  first  element  o.re  zeros,  the  determinant  reduces  to  ai^a^. 
Consider  the  determinant 


«i    0 

0 

0 

h,  h 

h 

b. 

Ci      c. 

Ci 

c. 

d,    d. 

d. 

d. 

Every  term  of  A  contains  one,  and  only  one,  element 
from  the  first  row ;  and  all  the  terms  that  do  not  contain 
«i  contain  one  of  the  zeros,  and  therefore  vanish.  The  terms 
that  contain  a^  contain  no  other  element  from  the  first  row 
or  column,  and,  consequently,  contain  one,  and  only  one, 
element  from  each  row  and  column  of  the  determinant 


,  or  Aa^. 


Hence,  disregarding  the  sign,  each  term  of  A  consists  of 
«!  multiplied  into  a  term  of  A^^. 

Take  any  particular  term  of  A,  as  a-})^c^d^ ;  the  sign  is 
fixed  (§  519,  Rule  I.)  by  the  number  of  inversions  in  the 


h. 

h 

J. 

Ci 

c% 

c» 

ck 

d. 

d. 

DETERMINANTS.  611 

series  14  3  2;  the  sign  of  the  term  h^c^di  of  A^^  is  fixed 
by  the  number  of  inversions  in  the  series  4  3  2.  Adding 
«!  makes  no  new  inversions  among  either  the  letters  or  the 
subscripts.  Consequently  the  sign  of  the  term  in  A  is  the 
same  as  the  sign  of  the  term  in  aiA^^. 

Since  this  is  true  of  every  term  of  A,  we  have 

A  =  aAa^' 
Similarly  for  any  determinant  of  like  form. 

533.  Terms  containing  an  Element.  From  §  532  it  appears 
that  the  sum  of  the  terms  which  contain  ai  may  be  written 
«iAaj.  For,  no  one  of  the  terms  which  contain  a^  can  con- 
tain any  one  of  the  elements  aj,  «3,  cia, and  these  terms 

are  therefore  unchanged  if  for  a2,  a-i,  a^,  in  the  given 

determinant  we  put  zeros. 

If  we  carry  the  second  column  over  the  first,  the  deter- 
minant is  changed  to  —  A.  By  §  532  the  sum  of  the  terms 
of  —  A  which  contain  aj  is  ajAa.^,  and  the  sum  of  the  corre- 
sponding terms  of  A  is  therefore  —  a^^a.^. 

In  general,  for  the  element  of  the  ^th  row  and  qth.  col- 
umn, we  shall  have  to  carry  the  pth  row  over  p  —  1  rows, 
and  the  ^th  column  over  q  —  1  columns  in  order  to  bring 
the  element  in  question  to  the  first  row  and  first  column. 
The  new  determinant  is  A  if  p-{-  q  —  2is  even,  and  is  —  A 
if  ^  +  5  —  2  is  odd  (§  524).  Consequently,  the  sum  of  the 
terms  of  A  which  contain  the  element  of  the  pth.  row  and 
^'th  column  is  the  product  of  that  element  by  its  minor ; 
the  sign  being  +  if  ^  -f  5^  is  even,  and  —  if  ^  +  3'  is  odd. 


the  sum  of  the  terms  which  contain 


Thus,  in 

a., 

«3 

a. 

the  sum 

d. 

C3 
^3 

C4 
d. 

C3  is  C3A 

Here  p  = 

3,? 

=  3, 

and 

p  + 

q  is  even 

512 


ALGEBRA. 


534.  Oo-factors.  Since  every  term  contains  one  element 
from  each  row  and  column,  if  we  add  together  the  sum  of 
the  terms  containing  ai,  the  sum  of  the  terms  containing  a.^, 
and  so  on,  we  shall  obtain  the  whole  expansion  of  the  given 
determinant. 

Thus,  in  the  determinant  \ai  h^  c^  d^], 

A  =  ttiAa,  —  aAa^  4-  a-Aa^  —  aAai\ 
The  expressions  A^^,   —  A^^,  A^g,  —  A^^  are  called  the 
co-factors  of  the  several  elements  a^,  a^,  as,  a^,  and  are  gen- 
erally represented  by  A^,  A^,  A^,  A^. 

Hence,  in  the  case  of  |ai  ^2  ^3  ^^l,  we  may  write 
A  =  aiAi  +  a^A^  +  a^A^  +  ^4744, 

=  a,A,  +  b,B,  +  c,C,-i-d,I)„ 
and  so  on.     Similarly  for  any  determinant. 

535.  Theorem.  If  the  elements  in  any  roiv  are  multiplied 
by  the  co-factors  of  the  corresponding  elements  in  another 
roiv,  the  sum  of  the  products  vanishes. 

Thus,  in  the  determinant  \ai  b.^  c^  dA 


b,B,-i-b,B,  +  b,B,  +  b,B, 


No  one  of  the  co-factors  Bi,  B^,  B^,  B^,  contains  any  of 
the  elements  b^,  b^,  b^,  b^.  These  co-factors  will,  conse- 
quently, be  unaffected  if  in  the  above  identity  we  change 
^1,  ^2,  ^3,  b^  to  «!,  «2,  <^3,  <^4-     This  gives 


rt, 

rt. 

ff-s 

a, 

K 

h 

h. 

h, 

C\ 

Ci 

Ct 

Oi 

d. 

d. 

d. 

d, 

ctiBi  -j-  ajBi  -f  ^3-^3  +  ctiBf, 
Similarly  for  any  other  case. 


«1 

a. 

«3 

a, 

ai 

a. 

a. 

a. 

Ci 

Ci 

C3 

c. 

d, 

d. 

d. 

d. 

DETERMINANTS. 


513 


536.  Evaluation  of  Determinants.  By  using  §  517,  §  519, 
and  §  534  we  can  readily  obtain  the  value  of  any  numer- 
ical determinant. 


Ex.  Evaluate 


From  the  first  row  subtract  3  times  the  second,  from  the  third 
twice  the  second,  from  the  fourth  4  times  the  second.     The  result  is 

0      -8      -2  -2 
13          2  1 

0-5-1  1 

0      -9      -6  -1 


which,  by  ^  534,  reduces  to 

- 

-8     -2     -2 
-5     -1         1 
-9     -6     -1 

or 

8  2         2 
5      1      -1 

9  6         1 

=  70  (\  521). 

537.   Simultaneous  Equations.     Consider  the  simultaneous 
equations 

Write  the  determinant 

«i     bi     Ci 

a^     b-i     C2  ,  and  let  Ai,  A^^  B^, 

1 

«3     ^3     c-i 

^2.  etc.,  be  the  co-factors  in  this  determinant. 

Multiply  the  first  equation  by  A^,  the  second  by  A^,  the 
third  by  ^3,  and  add.     The  result  is 

since  (§  535),        b^A^  +  b^A^  +  ^3^3  =  0, 
and  CxAx  +  c^A-i  -\-  c^A^  —  0. 


514 


ALGEBRA. 


Hence  (§  534),  we  see  that 


Als 


y 


^1        ^1        Cx 
1^1         (^2         Ci 

h     h     (?3 


or  X 


I  «1  ^2  O3  I" 


Q?i  ^'2  h  I 


^1  5,  ^3  I 


Similarly  for  any  set  of  simultaneous  equations  of  the 
first  degree. 

Exercise  147. 

1.  In  the  determinant  \ai  h.^  c^  o?4 1  write  the  co-factors 
of  a-i,  hi,  bi,  Ci,  <?4,  di,  d^. 

2.  Express  as  a  single  determinant 


e    f  g 

h     e    g 

^    9    I 

h     f     e 

f  h  k 

+ 

c    f    Ic 

+ 

c     h     h 

+ 

c     h     f 

9     ^     I 

d    g     I 

d     I    Tc 

dig 

Expand  : 


a 

h 

b 

a 

h 

a 

a 

b 

a 

a 

b 

b 

.    4. 

0 

a 

b 

b 

0 

d 

d 

d 

a 

0 

a 

a 

.    5. 

b 

b 

0 

b 

c 

c 

c 

0 

1 

a 

a 

a 

1 

b 

a 

a 

1 

a 

b 

a 

1 

a 

a 

b 

6. 


Find  the  value  of: 

3  2  2  2 

2  3  2  2 

2  2  3  2 

2  2  2  3 


3 

2 

1      4 

15 

29 

2    14 

.    8. 

16 

19 

3    17 

33 

39 

8    38 

Solve  the  equations : 

3a;-42/  +  2z=      l^ 
9.    2x\-2>y-^z  =  -l  ,' 


2  13  4 
7    4    5  9 

3  3    6  2 
17    7  5 


4a:-7y+    2  =  16- 

10.    3a; -f    y-2z-=l() 

5^_6y-32  =  10 


CHAPTER  XXXVIIL 


COMPLEX   NUMBERS. 

538.   Eepresentation  of  Pure  Imaginaries.    Represent  V—  1 
by  i.     Assuming  the  commutative  and  associative  laws, 

i  X  b,  that  is,  (+  i)  h  —  -{-  hi  ; 


§234. 


ixix  i  Xb  =  i^b  =  (~  i)b  =  —  bi 
ixiX  ixixb  =  i*b  =  (-f-  l)b  =  +  b. 

Hence,  two  multiplications  by  i  change  b  to  —  b  ;  that 
is,  two  multiplications  by  i  turn  the  line  represented  by  b 
through  180° ;  and  four  multiplications  by  i  turn  the  line 
represented  by  b  through  360°  ;  and  so  on. 

We  may  therefore  consistently  assume  that  one  multipli- 
cation by  i  turns  the  representative  line  through  90°,  three 
multiplications  by  i  through  270°,  and  so  on. 


X'- 


-b 


+-- 


It 


+  ?> 


-X 


H  then  we  draw  through  0,  the  zero  point  of  the  line  of 
real  numbers,  a  line  YY^  perpendicular  to  XX\  all  pure 


516  ALGEBRA. 

imaginaries  will  be  represented  by  lengths    on   this  line, 
just  as  all  real  numbers  are  represented  by  lengths  on  XX\ 

539.  Vectors.  A  directed  straight  line  of  definite  length 
is  called  a  vector. 

Two  parallel  vectors  which  have  the  same  length,  and 
extend  in  the  same  direction,  are  said  to  be  equal  vectors. 

540.  Vector  Addition.  To  add  a  vector  CD  to  a  vector 
AB,  we  place  C  on  B,  keeping 
CD  parallel  to  its  original  posi- 
tion, and  draw  AD.     Then, 

AD^AB-^^BD 

=  AB-\-CD.         (1) 

The  addition  here  meant  by 
/-^^  the  sign  +  is  not  addition  of 
numbers,  but  addition  of  vectors,  generally  called  geometric 
addition.     It  is  identical  with  the  composition  of  forces. 

From  the  dotted  lines  in  the  figure,  and  the  known  prop- 
erties of  a  parallelogram,  it  is  seen  that 

AD=CD-\-AB.  (2) 

From  (1)  and  (2),    AB -\- CD  =  CD -\-  AB. 

Consequently,  vector  addition  is  commutative  (§  31).  It 
is  easily  seen  that  it  is  also  associative  (§  32). 

541.  Complex  Numbers.  A  complex  number  in  general 
consists  of  a  real  part  and  an  imaginary  part,  and  may  be 
written  (§  235)  in  the  typical  form  a  +  hi,  where  a  and  b 
are  both  real. 

If  we  understand  the  sign  -f-  fo  indicate  geometric  addi- 
tion, we  shall  obtain  the  vector  which  represents  a  +  bi  as 
follows : 


COMPLEX    NUMBERS. 


517 


X-t 


Lay  off  a  on  the  axis  of  reals  from  0  to  M.  From  M 
draw  the  vector  MP  to  represent  hi.  Then  the  vector  OP 
is  the  geometric  sum  of  the  vectors  OM  and  MP,  and 
represents  the  complex 
number  a  +  bi. 

In  the  figure,  the  vec- 
tors OP,  OQ,  'OR,  OS, 
respectively,  represent  the 
complex  numbers  6  +  4z, 
-64-5z, -5-3z,  3  — 5z. 

In  the  complex  number 
a  +  bi,  a  and  5i  are  repre- 
sented by  vectors.  Now 
vector  addition  is  commu- 
tative.    Consequently,  a -\- hi  =  bi -{-  a. 

This  is  also  evident  from  the  figure. 

The  expression  a-\-bi  is  the  general  expression  for  all 
numbers.  This  expression  is  zero  when  a  =  0  and  i  =  0  ; 
is  a  real  number  when  5  =  0 ;  a  pure  imaginary  when  a  =  0  ; 
a  complex  number  when  a  and  b  both  differ  from  0. 

542.  Addition  of  Complex  Numbers.  Let  a  -\-  bi  and  a'  +  h^i 
be  two  complex  numbers.  Their  sum,  a -{- hi -\- a^  -\r  hH, 
may  by  the  commutative  law  be  written  a -\-  a'  -jr  (b -}-  b')i. 

Let  OJ  and  OB  be  the 
representative  vectors  of 
a-{-hi  2iud  a'-\-h'i.   Take 

AC-=  and  II  to  OB; 
then,  OC=OA-\-OP. 

Draw  the  other  lines 
in  the  figure. 

Then, 
0^=  0P+  FH 

=  OF-\-OE=a-\-a\  o 


518 


ALGEBRA. 


and  HC^FA-^KC 

•••  0^=a  +  a'  +  (^  +  ^')^'^ 
But  0C-- 


{a  +  hi)  +  (a'  +  Vi), 
OA  +  OB. 


Consequently,  the  geometric  sum  of  the  vectors  of  two  com- 
plex numbers  is  the  vector  of  their  suvi. 

Since  vector  addition  is  commutative,  it  follows  that  the 
addition  of  complex  numbers  is  commutative. 

The  sum  of  two  complex  numbers  is  the  geometric  sum 
of  the  sum  of  the  real  parts  and  the  sum  of  the  imaginary 
parts  of  the  two  numbers. 

The  preceding  may  be  made  clearer  by  a  particular  example. 
Find  the  sum  of  2  +  3  i  and  —  4  +  i. 

2  +  3  i  =  OM  and  —  4  +  i  =  0M\     If  now  we  proceed  from  M, 
Y  the  extremity  of  OM,  in  the 

direction  of  OM^  as  far  as 
the  absolute  value  of  0M\ 
we  reach  the  point  M^^. 

Hence,  0M'^  =  -2+^i, 
the  sum  of  the  two  given 
complex  numbers. 

The  same  result  is  reached 
if  we  first  find  the  value  of 
X'  A'  A''  0  AX    2  +  (-4)  =  -2.     That  is,  if 

we  count  from  0  two  real  units  to  A^^,  and  add  to  this  sum  3i  +  i  =  ii; 
that  is,  count  four  imaginary  units  from  A^^  on  the  perpendicular 
A'^M^^. 

543.  Modulus  and  Amplitude.  Any  complex  number, 
a  +  bi,  can  be  written  in  the  form 

VVa'  +  b''      Vn'  +  b'  ) 


The  expressions 


an( 


VoH^  Va'  +  h^ 


may  be  taken 


Complex  numbers. 


519 


as  the  sine  and  cosine  of  some  angle  <^,  since  they  satisfy 
the  equation 

cos^  <^  +  sin^  <^  —  1. 


If  we  put  r  =  'yjd'-  +  6^  the  complex   number   may  be 
written 

r  (cos  <^  +  *  sin  ^\ 


Since  r  —  ^d^  +  6^  the  sign  of  r  is  indeterminate.  "VVe 
shall,  however,  take  r  always  positive. 

The  positive  num^r  r  is  called  the  modulus,  the  angle  <^ 
the  ^Slp^tMfel'^ortELe  complex  number  a  +  bi. 

Let  OP  be  the  representative  vector  of  a  +  ^^-  Since  r 
is  the  positive  value  of  Va^M^,  it  is  evident  that  r  is  the 
length  of  OP.     Since 

a  a      OM 

cos<^ 


Va'  +  h' 


OP 


MP 
OP^ 


M 


and 

h  h 

sin  d>  =  — :r=^=:  =  - 
Va'^  +  b'      r 

it  follows  that  <^  is  the  angle  MOP. 

The  above  is  easily  seen  to  hold  true  when  a  and  b  are 
one  or  both  negative. 

The  modulus  of  a  real  number  is  its  absolute  value ;  the 
amplitude  is  0  if  the  number  is  positive,  180°  if  the  num- 
ber is  negative. 

The  modulus  of  a  pure  imaginary  bi  is  b  ;  the  amplitude 
is  90°  if  b  is  positive,  270°  if  b  is  negative. 

544.  Since  the  sum  of  the  lengths  of  two  sides  of  a 
triangle  is  greater  than  the  length  of  the  third  side,  it 
follows  from  §§  540,  542,  that  the  modulus  of  the  sum  of 
two  complex  num,bers  is  less  than  the  sum  of  the  moduli. 


520 


ALGEBRA. 


In  one  case,  however,  that  in  which  the  representative 
vectors  are  collinear,  the  modulus  of  the  sum  is  equal  to 
the  sum  of  the  moduli. 

545.  Multiplication  by  Eeal  Numbers.  Let  a  +  hi  be  any- 
complex  number.  If  the  representative  vector  be  multi- 
plied by  any  real  number  c,  it  is  easily  seen  from  a  figure 
that  the  product  is  ca  +  chi. 

Therefore,  c{a-{-  hi)  =  ca~\-  chi. 

It  follows  that  the  multiplication  of  a  complex  number 
by  a  real  number  is  distrihutive. 


Ex.  To  multiply  -  2  +  t  by  3  :  Take 
OA  =  —  2  on  OX^,  and  erect  at  A  the  per- 
pendicular AM=  1 .  Then  0M=  —  2  +  i; 
and,  by  taking  OM  three  times,  the 
result  is  OM''  =  —  6  +  3t,  the  product 
of  (-  2  +  i)  by  3. 


546.  Multiplication  by  Pure  Imaginaries.  We  have  seen 
(§  538)  that  multiplying  a  real  number,  or  a  pure  imaginary, 
by  i  turns  that  number  through  90°.  Let  us  consider  the 
effect  of  multiplying  a  complex  number  by  i. 

By  the  commutative,  associative,  and  distributive  laws, 

i  X  r  (cos  (f> -{- i  sin  <ji)  =  r  (i  cos  <^  —  sin  <f>) 

=  r(—  sin  <}>-}- i  cos  <f>) 
by  Trigonometry,  =  r  [cos  (90°  +  <^)  +  ^■  sin  (90°  +  <^)]. 

Here,  also,  the  effect  of  multiplying  by  i  is  to  increase 
<^  to  <^  +  90° ;  that  is,  to  turn  the  representative  vector 
in  the  positive  direction  through  an  angle  of  90°. 

The  effect  of  multiplying  by  a  pure  imaginary  hi  will  be 
to  turn  the  complex  number  through  a  positive  angle  of 
90°,  and  also  to  multiply  the  modulus  by  h. 


COMPLEX   NUMBERS.  521 

547.   Multiplication  by  a  Complex  Number.     We  come  now 

to  the  general  problem  of  the  multiplication  of  one  complex 
number  by  another.  This  includes  all  other  cases  as  par- 
ticular cases. 

Let  r  (cos  <f> -}- i  sin  <^)  and  r'  (cos  <j>'  +  i  sin  <^')  be  two 
complex  numbers.  By  actual  multiplication  their  product 
is 

rr'  [cos  (j)  cos  <^'  —  sin  <^  sin  <f>'  +  i  (sin  «^  cos  <^'  +  cos  <}>  sin  <^')}. 

By  Trigonometry,  this  may  be  written 

n-'  [cos  (<^  +  </)')  +  ^  si"  (</>  +  </»')]• 

Therefore,  the  modulus  of  the  product  of  twojCpmplex 
numbers  is  the  product  of  their  moduli ;  and  the  aim^lAUi^ 
of  the  product  is  the  sumfi  of  their  ^r^^itud^sT''"^ 

Consequently,  the  effect  of  multiplying  one  complex 
number  by  another  is  to  w.ultiply  the  modulus  of  the  first  hy 
the  modulus  of  the  second;  and  to  turn  the  representative 
vector  of  the  first  through  the  amplitude  of  the  second. 


548.   Division  by  a  Complex  Number.     The  quotient 

r(cos  <^  -f-^'sin  </)) 
r'(cos<^'  +  ^sin  ^') 

becomes,    multiplying    numerator    and    denominator    by 
cos  </>'  —  i  sin  <^', 

^[c■os(<^~<^')  +  ^•sin(</»-<^')]. 

Consequently,  the  modulus  of  the  quotient  is  obtained  by 
dividing  the  modulus  of  the  dividend  by  that  of  the  divi- 
sor ;  and  the  amplitude  of  the  quotient,  by  subtracting  the 
amplitude  of  the  divisor  from  that  of  the  dividend. 


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